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  • zoj 3396 Conference Call

    2014-05-27 19:54:23
    Good News! Utopia Polycom begins to offer a whole new service, Conference Call. This new service enables three users to make phone calls at the same time and talk to each other. It's very useful w
    
    Good News! Utopia Polycom begins to offer a whole new service, Conference Call. This new service enables three users to make phone calls at the same time and talk to each other. It's very useful when several people are discussing about one shared topic. For example, Tom, Rosemary and you plan to go picnic tomorrow. You just need to make one conference call, so that you can discuss with Tom and Rosemary simultaneously. It's extremely convenient, yes?

    However, it has a lot of technological problems to start this service. It took engineers of Utopia Polycom five years to conquer all these problems. One of the biggest problem is how to make the communication line connected. To solve this problem, they have constructed a powerful net of transfer stations. Every single phone is connected to its nearest transfer station, some transfer stations are connected to each other by wires. When to make a conference call, signal will start from the dialling telphone. The signal will be transferred through transfer stations and finally reach the other two phones.

    Since the distances between each two transfer stations are various, the cost for the signal to go through the wires is different as well. Assume there aren telphones(numbered from 1 to n) and m(numbered from1 to m) transfer stations in this communication net. The total cost for a line is the sum of the costs of the wires in the line. Our problems remains, what is the minimal total cost to make a conference call among three telphonea, b and c. As showing below, the minimal total cost to make a conference call among telphone1, 2, 3 is 12.

    Input

    There are multiple cases. For each test case, the first line contain three positive integers,n (n ≤ 10000), m (m ≤ 500), l (l ≤ 10000).l means the number of wires. Then follows n lines. Each line contains one integerti (1 ≤ tim), which means the number of transfer station No.i telphone is connected to. Then followsl lines, each line contains three integers a, b, C (1 ≤C ≤ 500), which means a and b transfer stations are connected to each other by a wire with the amount of costC. If a pair (a, b) doesn't appear in these l lines, that means tranfer stationsa, b are not connected.

    The next line is a single integer q (q ≤ 50). Then follow q lines. Each line contains three integers x, y, z. You are supposed to calculate the minimal total cost to make a conference call among telphonex, y, z. Each test case is seperated by a blank line. Proceed to the end of the file.

    Output

    For each test case i, print case number in the form "Case #i" in one single line. Then for each inquiryj, print one line in the form "Line j: " and if the conference call can be made and the minimal total cost isMin, print the sentense "The minimum cost for this line is Min." else you should print "Impossible to connect!". Look at the samples for details.

    Sample Input

    3 5 5
    2
    1
    5
    1 2 6
    2 3 1
    3 4 2
    4 5 3
    1 5 12
    1
    1 2 3
    
    3 3 1
    1
    2
    3
    2 3 4
    1
    1 2 3
    

    Sample Output

    Case #1
    Line 1: The minimum cost for this line is 12.
    Case #2
    Line 1: Impossible to connect!
     
    
    题意:现在某公司推出一项新项目,支持三个人之间的电话通化,每台电话连接一个最近的中转站,然而要进行通话,是需要在中转站上传输信号,传输的距离越长费用就越高,求出给定的三台电话的通话所需的最短路程。
    思路:使用floyd算法就出每两个城市之间的最短路径,然后依据这些路径求出三台电话之间的最短路径。求的时候还是有一个大坑的,不能只枚举给定三个城市之间的最短路。例如:看下图,假设求A,B,C三点的最短路,如果只涉及A,B,C三点的话,求出的路径显然不是最短的,假设以D为中心,求出来的路径肯定比刚才的路径要短。
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #define INF 100000000
    using namespace std;
    int root[10004];         //用来存储每台电话最近的中转站
    int f[504][504];
    int n,m,l;
    void init()
    {
        for(int i=0; i<=n; i++)
            for(int j=0; j<=i; j++)
            {
                if(i!=j) f[i][j]=f[j][i]=INF;
                else f[i][j]=0;
            }
    }
    
    void floyd()
    {
        for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(f[i][j]>f[i][k]+f[k][j])
            {
                f[i][j]=f[i][k]+f[k][j];
            }
        }
    }
    int main()
    {
        int ans=1;
        while(scanf("%d%d%d",&m,&n,&l)!=EOF)
        {
            init();
            for(int i=1;i<=m;i++)
            scanf("%d",&root[i]);
            int x,y,z;
            for(int i=0;i<l;i++)
            {
                scanf("%d%d%d",&x,&y,&z);
                if(f[x][y]>z) f[x][y]=f[y][x]=z;
            }
            floyd();
            scanf("%d",&l);
            printf("Case #%d\n",ans++);
            for(int i=1;i<=l;i++)
            {
                scanf("%d%d%d",&x,&y,&z);
                x=root[x];
                y=root[y];
                z=root[z];
                int sum=INF;
                for(int i=1;i<=n;i++)          //本题关键,很坑....
                {
                    if(sum>f[i][x]+f[i][y]+f[i][z])
                        sum=f[i][x]+f[i][y]+f[i][z];
                }
    
                printf("Line %d: ",i);
                if(sum>=INF) printf("Impossible to connect!\n");
                else printf("The minimum cost for this line is %d.\n",sum);
            }
        }
        return 0;
    }
    

    展开全文
  • ZOJ 3396 Conference Call

    千次阅读 2014-02-22 21:10:02
    Conference Call Time Limit: 2000MS   Memory Limit: 65536KB   64bit IO Format: %lld & %llu [Submit]  [Go Back] [Status]  Description Good News! Utopia Polycom ...


    Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

    []   [Go Back]   [Status]  

    Description

    Good News! Utopia Polycom begins to offer a whole new service, Conference Call. This new service enables three users to make phone calls at the same time and talk to each other. It's very useful when several people are discussing about one shared topic. For example, Tom, Rosemary and you plan to go picnic tomorrow. You just need to make one conference call, so that you can discuss with Tom and Rosemary simultaneously. It's extremely convenient, yes?

    However, it has a lot of technological problems to start this service. It took engineers of Utopia Polycom five years to conquer all these problems. One of the biggest problem is how to make the communication line connected. To solve this problem, they have constructed a powerful net of transfer stations. Every single phone is connected to its nearest transfer station, some transfer stations are connected to each other by wires. When to make a conference call, signal will start from the dialling telphone. The signal will be transferred through transfer stations and finally reach the other two phones.

    Since the distances between each two transfer stations are various, the cost for the signal to go through the wires is different as well. Assume there are n telphones(numbered from 1 to n) and m(numbered from1 to m) transfer stations in this communication net. The total cost for a line is the sum of the costs of the wires in the line. Our problems remains, what is the minimal total cost to make a conference call among three telphone ab and c. As showing below, the minimal total cost to make a conference call among telphone 123 is 12.

    Input

    There are multiple cases. For each test case, the first line contain three positive integers, n (n ≤ 10000), m (m ≤ 500), l (l ≤ 10000). l means the number of wires. Then follows n lines. Each line contains one integer ti (1 ≤ ti ≤ m), which means the number of transfer station No.i telphone is connected to. Then follows l lines, each line contains three integers abC (1 ≤ C ≤ 500), which means a and b transfer stations are connected to each other by a wire with the amount of cost C. If a pair (ab) doesn't appear in these l lines, that means tranfer stations ab are not connected.

    The next line is a single integer q (q ≤ 50). Then follow q lines. Each line contains three integers xyz. You are supposed to calculate the minimal total cost to make a conference call among telphone xyz. Each test case is seperated by a blank line. Proceed to the end of the file.

    Output

    For each test case i, print case number in the form "Case #i" in one single line. Then for each inquiry j, print one line in the form "Line j: " and if the conference call can be made and the minimal total cost isMin, print the sentense "The minimum cost for this line is Min." else you should print "Impossible to connect!". Look at the samples for details.

    Sample Input

    3 5 5
    2
    1
    5
    1 2 6
    2 3 1
    3 4 2
    4 5 3
    1 5 12
    1
    1 2 3
    
    3 3 1
    1
    2
    3
    2 3 4
    1
    1 2 3
    

    Sample Output

    Case #1
    Line 1: The minimum cost for this line is 12.
    Case #2
    Line 1: Impossible to connect!

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    
    using namespace std;
    
    const int INF=0x3f3f3f3f;
    
    int N,M,L,Q,telephone[11000],G[600][600],dist[600][600],vis[600];
    bool use[600];
    
    void dijkstra(int src)
    {
        dist[src][src]=0;
        memset(use,false,sizeof(use));
        for(int i=0;i<M;i++)
        {
            int mark=-1,dd=INF;
            for(int j=1;j<=M;j++)
            {
                if(use[j]) continue;
                if(dist[src][j] < dd)
                {
                    dd=dist[src][j]; mark=j;
                }
            }
    
            if(mark==-1) return;
    
            use[mark]=true;
            for(int j=1;j<=M;j++)
            {
                if(use[j]) continue;
                if(dist[src][j]>dist[src][mark]+G[mark][j])
                {
                    dist[src][j]=dist[src][mark]+G[mark][j];
                }
            }
        }
    }
    
    int main()
    {
         int cas=1;
    while(scanf("%d%d%d",&N,&M,&L)!=EOF)
    {
        printf("Case #%d\n",cas++);
        for(int i=1;i<=N;i++)   scanf("%d",telephone+i);
        memset(G,63,sizeof(G));
        memset(dist,63,sizeof(dist));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<L;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            G[a][b]=G[b][a]=c;
        }
        scanf("%d",&Q);
    
        for(int j=1;j<=Q;j++)
        {
            printf("Line %d: ",j);
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            x=telephone[x];
            y=telephone[y];
            z=telephone[z];
    
            if(!vis[x]) vis[x]=1,dijkstra(x);
            if(!vis[y]) vis[y]=1,dijkstra(y);
            if(!vis[z]) vis[z]=1,dijkstra(z);
    
            int ans=INF;
            for(int i=1;i<=M;i++)
            {
                ans=min(ans,dist[x][i]+dist[y][i]+dist[z][i]);
            }
    
            if(ans==INF) printf("Impossible to connect!\n");
            else printf("The minimum cost for this line is %d.\n",ans);
        }
    }
        return 0;
    }
    



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  • Conference Call Time Limit: 2 Seconds Memory Limit: 65536 KB ...Good News! Utopia Polycom begins to offer a whole new service, Conference Call. This new service enables three users to make pho
    Conference Call

    Time Limit: 2 Seconds       Memory Limit: 65536 KB

    Good News! Utopia Polycom begins to offer a whole new service, Conference Call. This new service enables three users to make phone calls at the same time and talk to each other. It's very useful when several people are discussing about one shared topic. For example, Tom, Rosemary and you plan to go picnic tomorrow. You just need to make one conference call, so that you can discuss with Tom and Rosemary simultaneously. It's extremely convenient, yes?

    However, it has a lot of technological problems to start this service. It took engineers of Utopia Polycom five years to conquer all these problems. One of the biggest problem is how to make the communication line connected. To solve this problem, they have constructed a powerful net of transfer stations. Every single phone is connected to its nearest transfer station, some transfer stations are connected to each other by wires. When to make a conference call, signal will start from the dialling telphone. The signal will be transferred through transfer stations and finally reach the other two phones.

    Since the distances between each two transfer stations are various, the cost for the signal to go through the wires is different as well. Assume there are n telphones(numbered from1 to n) and m(numbered from 1 to m) transfer stations in this communication net. The total cost for a line is the sum of the costs of the wires in the line. Our problems remains, what is the minimal total cost to make a conference call among three telphone ab and c. As showing below, the minimal total cost to make a conference call among telphone 123 is 12.

    Input

    There are multiple cases. For each test case, the first line contain three positive integers, n (n ≤ 10000), m (m ≤ 500), l (l ≤ 10000). l means the number of wires. Then follows n lines. Each line contains one integer ti (1 ≤ ti ≤ m), which means the number of transfer station No.i telphone is connected to. Then follows l lines, each line contains three integers abC(1 ≤ C ≤ 500), which means a and b transfer stations are connected to each other by a wire with the amount of cost C. If a pair (ab) doesn't appear in these l lines, that means tranfer stations ab are not connected.

    The next line is a single integer q (q ≤ 50). Then follow q lines. Each line contains three integers xyz. You are supposed to calculate the minimal total cost to make a conference call among telphone xyz. Each test case is seperated by a blank line. Proceed to the end of the file.

    Output

    For each test case i, print case number in the form "Case #i" in one single line. Then for each inquiry j, print one line in the form "Line j: " and if the conference call can be made and the minimal total cost is Min, print the sentense "The minimum cost for this line is Min." else you should print "Impossible to connect!". Look at the samples for details.

    Sample Input

    3 5 5
    2
    1
    5
    1 2 6
    2 3 1
    3 4 2
    4 5 3
    1 5 12
    1
    1 2 3
    
    3 3 1
    1
    2
    3
    2 3 4
    1
    1 2 3
    

    Sample Output

    Case #1
    Line 1: The minimum cost for this line is 12.
    Case #2
    Line 1: Impossible to connect!
    

    Author:  ZHU, Yuke

    Contest: ZOJ Monthly, September 2010


    对三个点跑一遍最短路,然后枚举m个点作为中间点,取最小值


    #include <cstdlib>
    #include <cctype>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <string>
    #include <iostream>
    #include <sstream>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <fstream>
    #include <numeric>
    #include <iomanip>
    #include <bitset>
    #include <list>
    #include <stdexcept>
    #include <functional>
    #include <utility>
    #include <ctime>
    
    using namespace std;
    
    #define PB push_back
    #define MP make_pair
    #define REP(i,n) for(int i=0;i<(n);++i)
    #define FOR(i,l,h) for(int i=(l);i<=(h);++i)
    #define DWN(i,h,l) for(int i=(h);i>=(l);--i)
    #define CLR(vis) memset(vis,0,sizeof(vis))
    #define MST(vis,pos) memset(vis,pos,sizeof(vis))
    #define MAX3(a,b,c) max(a,max(b,c))
    #define MAX4(a,b,c,d) max(max(a,b),max(c,d))
    #define MIN3(a,b,c) min(a,min(b,c))
    #define MIN4(a,b,c,d) min(min(a,b),min(c,d))
    #define PI acos(-1.0)
    #define INF 0x7FFFFFFF
    #define LINF 1000000000000000000LL
    #define eps 1e-8
    
    
    const int maxn=555;
    
    int pos[maxn*20];
    int mp[maxn][maxn],dis[maxn][maxn];
    int flag[maxn];
    int vis[maxn];
    
    int n,m,l;
    
    void init()
    {
        CLR(flag);
        REP(i,maxn)
        {
            REP(j,maxn)
             mp[i][j]=-1,dis[i][j]=-1;
            mp[i][i]=0,dis[i][i]=0;
        }
    }
    
    
    void dijkstra(int from,int n)
    {
        int s=from;
        int g,res;
        CLR(vis);
        vis[s]=1;
        REP(i,n)
        {
            g=INF;
            FOR(j,1,n)
            {
                if(!vis[j])
                {
                    if(mp[s][j]!=-1 && dis[from][j]>dis[from][s]+mp[s][j] )
                        dis[from][j]=dis[from][s]+mp[s][j];
                    else
                      if(mp[s][j]!=-1 && dis[from][j]==-1 )
                        dis[from][j]=dis[from][s]+mp[s][j];
                    if(g>dis[from][j]&&dis[from][j]!=-1 )
                    {
                        g=dis[from][j];
                        res=j;
                    }
                }
            }
            if(g==INF)
                break;
            s=res;
            vis[s]=1;
        }
    }
    
    
    bool check(int a,int b,int c )
    {
        if(dis[a][b]==-1 || dis[a][c]==-1)
            return false;
        return true;
    }
    
    
    int main()
    {
        int cas=1;
        while(scanf("%d%d%d",&n,&m,&l)!=EOF)
        {
            init();
            FOR(i,1,n)
              scanf("%d",&pos[i]);
            int u,v,w;
            REP(i,l)
            {
                scanf("%d%d%d",&u,&v,&w);
                mp[u][v]=mp[v][u]=w;
            }
            int q;
            scanf("%d",&q);
            printf("Case #%d\n",cas++);
            int cnt=1;
    
            int a,b,c;
    
            while(q--)
            {
                scanf("%d%d%d",&a,&b,&c);
                a=pos[a],b=pos[b],c=pos[c];
                if(!flag[a])
                    flag[a]=1,dijkstra(a,m);
                if(!flag[b])
                    flag[b]=1,dijkstra(b,m);
                if(!flag[c])
                    flag[c]=1,dijkstra(c,m);
                if(check(a,b,c))
                {
                    v=INF;
                    FOR(i,1,m)
                      if(v>dis[a][i]+dis[b][i]+dis[c][i] )
                            v=dis[a][i]+dis[b][i]+dis[c][i];
                    printf("Line %d: The minimum cost for this line is %d.\n",cnt++,v);
                }
                else
                    printf( "Line %d: Impossible to connect!\n",cnt++);
    
            }
    
        }
    }
    



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  • Home Articles ...News Job Board About Home Articles NoPHP: A Conference Website in XForms Share on Facebook Tweet Google+ LinkedIn Email Print NoPHP: A Conference Website in XFor...

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    NoPHP: A Conference Website in XForms

    July 31, 2019

    Steven Pemberton
    xforms
    Steven Pemberton continues his series on XForms by constructing a conference website.
    Contents

    Introduction
    The Data
    The Talks
    The Conference Program
    The Program
    Dates and Times
    Error checking
    

    Introduction

    Consider this conference program overview:

    A Conference Overview Page

    If you look at the code, you can see that the page contains all the content, including abstracts for the talks, with the use of several scripts to handle pop ups and so on. The abstracts are initially hidden, but clicking on an “Abstract” button pops up the abstract for a talk:

    Talk abstract pop-up

    Let’s do this with XForms.
    The Data

    Firstly, we don’t want the data about the talks in the page. The talks will be stored in two files, the tutorials:

    XProc 3.0 Achim Berndzen Norman Walsh CSS Paged Media Tony Graham Declarative Applications with XForms Steven Pemberton

    and similar for the talks, but with abstracts:

    <Angle-brackets/> on the Branch Line - (model) railway engineering with XML technologies

    [Elided]

    John Lumley (jwLResearch)* Documenting XML Structures

    [Elided]

    Erik Siegel (Xatapult)* etc.

    We create two instances to store that data:

    Since this is to be generic, so that it can also be used in future years, we’ve added the date to the top element of the tutorials data. Now we can start displaying. Starting with the tutorials, we’ll begin with the prelude. The group element sets the context for the enclosed bit so that all enclosed references will by default be to data in the tutorials instance. Since we don’t know how many tutorials there will be each year we let XForms do the work, by outputting a count of the talks.

    The preconference day, , will comprise of tutorial sessions by some of the world's foremost experts in their respective fields:

    Now we can output the list of tutorials. Note that it deals with more than one author:

    which, with suitable CSS, looks like this:

    Source
    The Talks

    These are treated in the same way, except we now want to add a button to display the abstract. It starts the same:

    Then comes the button to show the abstract:

    Attaching the trigger like this to the abstract (ref=“abstract”) means that if there is no abstract, there’ll be no button (we’ve removed the abstract from one of the talks below to show this working).

    When the trigger is clicked on, or otherwise activated, a DOMActivate event is sent. This causes the toggle element to be processed.

    The toggle element causes a case of a switch element to be activated. Switch has a number of cases, only one of which is visible at a time. Initially by default the first is visible, and a different one can be activated with a toggle. In this case, we have a switch with two cases, one which shows nothing, and the other that shows the abstract:

    ... this is where we display the abstract ...

    To display the abstract, we repeat over the paragraphs of the data. This is preceded by a trigger to hide the abstract again by activating the other case:

    This looks like this (try it):

    Source

    This doesn’t pop up the abstract, but just makes it visible. Should popping up be essential to the user experience, then we just change the CSS for that. You really don’t need Javascript for these purposes:

    Source

    If your complaint is that it is formatted differently to the original, well, we just change the CSS for that as well (this isn’t a CSS tutorial, but if you need a hint, we’ve used display: inline-block).

    Source

    Finally, if a popping up animation is your thing, you can add that with a bit of CSS as well. (Look up CSS animations to discover how this was done.)

    Source
    The Conference Program

    Later, the conference changed the page to include the timings. Let’s adapt.

    The Conference Program
    The Program

    Our options here are to reorder and add times to the talks document we created, or to create a separate program document.

    Let’s do that.

    The program consists of a number of days, and each day consists of a number of slots. A slot has a start and end time, and a title:

    Registration Declarative Applications with XForms Break XProc 3.0 Lunch XProc 3.0 (cont'd) Break CSS Paged Media Registration Everyone Knows What a Dragon Looks Like

    etc.

    To display this we repeat over the days, and for each day, repeat over the slots:

    Here’s what this looks like:

    Source

    This of course is still missing the authors and the abstract. We get those from the talks instance by finding the talk whose title is the same as the entry in the slot, and then output the authors and abstract (in the same way as earlier):

    ... etc ...

    If no talk matches the condition, no talk will be selected, and so neither author nor abstract will be output. (By the way, we’ve merged the tutorials instance, and the talks instance for simplicity in this version, and used a different transition in the CSS for the abstracts).

    Here’s what it looks like:

    Source
    Dates and Times

    Since the data contains times, an extra feature that we can add is to show which items are in the past, which in the future, and which is happening now.

    For instance on Saturday at 11:30, the display could look like this:

    The conference program, with items displayed vaguely in the past, and the current item emphasised.

    To achieve this, we only have to compare the time for the slot with the time now: if the slot end time is less than the current time, the slot is in the past, if the start time is less than the current time, but the end time isn’t, then the slot is now active, and otherwise the slot is in the future.

    In essence, we set the class of the slot by calculating which of three it is, past, future, or now:

    That is the essence. Now the detail.

    To do the comparisons properly, we need both the time and the date. And because the computer displaying the page may be set to a different timezone, we need the timezone information as well.

    One approach would be to set all the start and end attributes in the program data file to include full date and time information:

    Registration

    and then display the times accordingly:

    Then the comparisons would be simple. We have the times in the timezone of the conference, but we need to convert them to the timezone of the computer displaying the page:

    class="{if(adjust-dateTime-to-timezone(@end) < instance(‘admin’)/now, ‘past’,
    if(adjust-dateTime-to-timezone(@start) < instance(‘admin’)/now, ‘now’,
    ‘future’))}">

    All that we then need to do is update the now value every minute. At startup we dispatch an event we shall call tick:

    and we catch it, update the now value, and dispatch another tick, with a delay of a minute (in milliseconds):

    That solution, though simple, does require a lot of fiddly work for the author of the data. The other option is to include the parts of the date and time in the data at suitable points:

    Registration

    (The timezone needs to be included per day, for the case that the conference straddles a change to or from daylight savings time.)

    In this case, the display of the times remain the same as the old method, but the comparisons have to be done differently. For each slot we have to construct the time to compare:

    concat(…/@date, ‘T’, @start, ‘:00’, …/@timezone)

    and again convert to the right timezone::

    adjust-dateTime-to-timezone(concat(…/@date, ‘T’, @start, ‘:00’, …/@timezone))

    Here it is working. I have faked the dates, so that the conference appears to start today, and the times are in your timezone, so that you can see it working.

    Source
    Error checking

    Suppose that there is a talk in the talks document that was forgotten in the program, or added twice to the program by mistake. To warn for that we repeat over the talk titles:

    and output those that don’t occur exactly once in the program:

    How do we find out how many times a title occurs in a program? We locate all slots in the program with that title:

    instance(‘program’)/day/slot[. = context()]

    and then count them:

    count(instance(‘program’)/day/slot[. = context()])

    That count should be 1.

    In full:

    If there are no missing or duplicated talks, nothing will show up.

    One other thing we can check for is that there are no unaccounted gaps between slots, by reporting all the slots whose start time doesn’t match the preceding slot’s end time:



    Source

    But if we can check that they abut, why not just assume they do? Each slot then only needs a start time, and unless there is an end time given, we can assume it ends at the start time of the next slot:

    Registration Declarative Applications with XForms Break XProc 3.0 Lunch XProc 3.0 (cont'd) Break CSS Paged Media ...

    In this version, instead of displaying the times with

    we display the end time if it is given, and otherwise the start time of the next slot:

    So here is the final version, again with faked dates and timezone, so that you can see it working, and a different CSS style for the current talk:

    Source

    It comprises 111 lines: 10 for the template, 26 lines of CSS, and 75 lines of XForms, of which 10 are there to support the CSS transition effects, and 2 for faking the dates.
    Article contents © 2019 Steven Pemberton
    Related links

    A News Carousel in XForms
    A Game in XForms
    A Clock in XForms
    A Calendar in XForms
    Viewing Data with XForms
    An Introduction to XForms
    

    © Textuality Services, Inc. except for those articles with named authors or copyright holders. All trademarks and registered trademarks appearing on XML.com are the property of their respective owners.

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