• Transportation
• transportation
• s start with transportation of physical resources. 1. Do we have a use case we can start working with? 2. If we want to transport 1000kg of apples, already packed in 50kg wooden crates. How do ...
• <div><p>Use https://github.com/cobalt-uoft/uoft-scrapers/issues/32 and https://github.com/cobalt-uoft/uoft-scrapers/issues/33 to create the Transportation API. It will provide data for UTM shuttles as...
• cityjson-transportation
• ai_transportation
• A transportation network consists of a central node and some other nodes directly or indirectly (through some intermediate nodes) connected to the central node, and also the transporting channels ...
• Heavy Transportation（POJ - 1797） Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 57578 Accepted: 14456 Description Background Hugo Heavy is happy. After the breakdown ...
Heavy Transportation（POJ - 1797）

Time Limit: 3000MS

Memory Limit: 30000K
Total Submissions: 57578

Accepted: 14456
Description

Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Scenario #1:
4

题意：求所有能从  1  --->  N  节点的所有路径，假如共m条路。  每一条路径中最小的那一段的最大值。

题解：

二分法：二分答案，再找1 到 N 的路径。当前二分值小的路径都不选用。若1到N存在路径。再把答案调大。若不能，调小答案。

AC代码：

//#include <bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>

using namespace std;
#define IO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mset(a, n) memset(a, n, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<int,PII> PPI;
typedef pair<ll, ll> PLL;
const int N = 1000+3;

int V,E;
struct node {
int v;
int d;
node(int v1,int d1): v(v1),d(d1){}
};
pair<int,int> P[N];
vector<node> G[N];
bool vis[N];
int dis[N];
int ans = 0;

bool spfa(const int lim)
{
memset(vis, false, sizeof(vis));
mset(dis,0x3f3f);
queue<int> Q;
Q.push(1);
dis[1] = 0;
vis[1] = true;

while(!Q.empty()){
int u = Q.front();Q.pop();

vis[u] = false;
for(int i=0;i<G[u].size();i++){
int v = G[u][i].v;
int val = G[u][i].d;
if(val<lim) continue;
if(dis[v] > (dis[u]+val)){
dis[v] = dis[u]+val;
if(!vis[v]) {
Q.push(v);
vis[v] = true;
}
}
}
}
if(dis[V] < 1061109567) return true;
else return false;
}

int main()
{
IO;
int tot;
cin>>tot;

for(int test = 1;test<=tot;test++){
cin>>V>>E;
for(int i=1;i<=V;i++) G[i].clear();
for(int i=0;i<E;i++){
int u,v,val;
cin>>u>>v>>val;
G[u].pb(node(v,val));
G[v].pb(node(u,val));
}

int l = 0, r = 1000001, mid;
while((r-l) >= 0){
mid = (r-l)/2+l;
if((spfa(mid))){
ans = mid;
l = mid+1;
}
else{
r = mid-1;
}
}
printf("Scenario #%d:\n",test);
printf("%d\n\n",ans);
}
return 0;
}

/*
10

3 3
1 3 4
2 3 1
1 2 2

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

5 7
1 2 2
2 5 2
1 3 4
1 4 7
3 4 1
2 3 1
3 5 6
*/


Dijkstra算法改编: dis[v] = max( min(dis[u], map[u][v]), dis[v] )  ,  其中  dis[u/v] 表示1 到 u/v 的路径中最小的那一段。

AC代码：

// 以后再补。
嘻嘻

最小生成树改编：   其实你可以把这道题看成是一道最大生成树。然后找到该树中最小的边。就是答案啦！

AC代码：

以后再补。。。

小博认为以上三种方法。第二种最好！有错的地方还请大家不吝赐教啊

展开全文
• For each case output the minimal total transportation cost. The minimal cost is guaranteed to be less than 231. Sample Input 3 1 2 3 0 1 1 0 2 1 3 1 3 3 0 1 3 0 2 4 2 1 2 0 1 1 Sample Output 1 3 ...
• The transportation problem is to minimize the cost of transporting good from m source nodes to n destination nodes through arcs. Arcs are directed routes from source to destination which have no ...
• <div><p>Public transportation infrastructure, such as stations, stops, routes, vending machines, and ticket validators, are tagged with <code>network</code> and <code>operator</code> tags. name-...
• Goods transportation time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There arencities located along the one-way r...

Goods transportation

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n cities located along the one-way road. Cities are numbered from 1 to n in the direction of the road.
The i-th city had produced pi units of goods. No more than si units of goods can be sold in the i-th city.
For each pair of cities i and j such that 1 ≤ i < j ≤ n you can no more than once transport no more than c units of goods from the city i to the city j. Note that goods can only be transported from a city with a lesser index to the city with a larger index. You can transport goods between cities in any order.
Determine the maximum number of produced goods that can be sold in total in all the cities after a sequence of transportations.

Input
The first line of the input contains two integers n and c (1 ≤ n ≤ 10 000, 0 ≤ c ≤ 109) — the number of cities and the maximum amount of goods for a single transportation.
The second line contains n integers pi (0 ≤ pi ≤ 109) — the number of units of goods that were produced in each city.
The third line of input contains n integers si (0 ≤ si ≤ 109) — the number of units of goods that can be sold in each city.

Output
Print the maximum total number of produced goods that can be sold in all cities after a sequence of transportations.

Examples

input
3 01 2 33 2 1

output
4

input
5 17 4 2 1 01 2 3 4 5

output
12

input
4 313 10 7 44 7 10 13

output
34分析：考虑最大流等于最小割，从小到大dp；　　　dp[i][j]表示前i个点有j个点在最小割点集里，　　　则dp[i][j]=min(dp[i-1][j-1]+s[i],dp[i-1][j]+j*c+p[i]);　　　dp[i-1][j-1]+s[i]表示i留在s-割的代价，dp[i-1][j]+j*c+p[i]表示i留在t-割，除了要花费p[i]外，还要花费j*c的代价；代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<ll,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
{
ll x=0;int f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,m,k,t;
ll p[maxn],s[maxn],ans[maxn],c;
int main()
{
int i,j;
scanf("%d%lld",&n,&c);
rep(i,1,n)scanf("%lld",&p[i]);
rep(i,1,n)scanf("%lld",&s[i]);
rep(i,1,n)
{
ans[i]=1e18;
for(j=i;j>=1;j--)
{
ans[j]=min(ans[j]+j*c+p[i],ans[j-1]+s[i]);
}
ans[0]+=p[i];
}
ll ret=1e18;
rep(i,0,n)ret=min(ret,ans[i]);
printf("%lld\n",ret);
//system("Pause");
return 0;
}

转载于:https://www.cnblogs.com/dyzll/p/5956661.html
展开全文
• transportation-homepage-design
• <p>There was a QGIS plugin developed by a French company that allowed transportation networks management, but development as ceased and there was no update for 3 years. It will work only with QGIS 1.0...
• Transportation You are making a program for a bus service. A bus can transport 50 passengers at once. Given the number of passengers waiting in the bus station as input, you need to calculate and ...
Transportation
You are making a program for a bus service.
A bus can transport 50 passengers at once.
Given the number of passengers waiting in the bus station as input, you need to calculate and output how many empty seats the last bus will have.
Sample Input:
126
Sample Output:
24
Explanation: The first bus will transport 50 passengers, leaving 126-50=76 in the station. The next one will leave 26 in the station, thus, the last bus will take all of the 26 passengers, having 50-26=24 seats left empty.
#include <iostream>
using namespace std;

int main() {
int pas;
cin >> pas;
pas%=50;
cout << 50-pas;
return 0;
}



展开全文
• uva 301 Transportation
原题：
Ruratania is just entering capitalism and is establishing new enterprising activities in many fields including transport. The transportation company TransRuratania is starting a new express train from city A to city B with several stops in the stations on the way. The stations are successively numbered, city A station has number 0, city B station number m. The company runs an experiment in order to improve passenger transportation capacity and thus to increase its earnings. The train has a maximum
capacity n passengers. The price of the train ticket is equal to the number of stops (stations) between the starting station and the destination station (including the destination station). Before the train starts its route from the city A, ticket orders are collected from all onroute stations. The ticket order from the station S means all reservations of tickets from S to a fixed destination station. In case the
company cannot accept all orders because of the passenger capacity limitations, its rejection policy is that it either completely accept or completely reject single orders from single stations. Write a program which for the given list of orders from single stations on the way from A to B determines the biggest possible total earning of the TransRuratania company. The earning from one accepted order is the product of the number of passengers included in the order and the price of their train tickets. The total earning is the sum of the earnings from all accepted orders.
Input
The input file is divided into blocks. The first line in each block contains three integers: passenger capacity n of the train, the number of the city B station and the number of ticket orders from all stations. The next lines contain the ticket orders. Each ticket order consists of three integers: starting station, destination station, number of passengers. In one block there can be maximum 22 orders. The number of the city B station will be at most 7. The block where all three numbers in the first line are equal to zero denotes the end of the input file.
Output
The output file consists of lines corresponding to the blocks of the input file except the terminating block. Each such line contains the biggest possible total earning.
Sample Input

10 3 4
0 2 1
1 3 5
1 2 7
2 3 10
10 5 4
3 5 10
2 4 9
0 2 5
2 5 8
0 0 0

Sample Output
19
34

中文：
首先给你3个数，表示这辆车的容量，车站数和订单数。
然后给你下面一堆订单，每个订单包含3个数，分别是起始站点，目的地站点，乘客数。
每个订单要取，要么不取，在保证车上人数不超过上限的情况下，最多能赚多少钱？每个乘客坐1站1元钱。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//fstream in,out;
struct node
{
int s,e,num;
};
node no[26];
int mark[100];
int cap,stat,ord,ans;
void update(int x)
{
for(int i=no[x].s;i<no[x].e;i++)
mark[i]+=no[x].num;
}
void recall(int x)
{
for(int i=no[x].s;i<no[x].e;i++)
mark[i]-=no[x].num;
}
bool judge(int x)
{
for(int i=no[x].s;i<no[x].e;i++)
{
if(mark[i]+no[x].num>cap)
return false;
}
return true;
}
void dfs(int x)
{
if(x==ord)
{
int tmp=accumulate(mark,mark+stat+1,0);
ans=max(ans,tmp);
}
else
{
if(judge(x))
{
update(x);
dfs(x+1);
recall(x);
}
dfs(x+1);
}
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>cap>>stat>>ord)
{
if(cap+stat+ord==0)
return 0;
ans=0;
memset(mark,0,sizeof(mark));
for(int i=0;i<ord;i++)
cin>>no[i].s>>no[i].e>>no[i].num;
dfs(0);
cout<<ans<<endl;
}
return 0;
}


解答：
直接用暴力dfs即可，由于车站数最多只有7个，所以每次更新的时候不需要使用区间数据结构，直接更新即可。
展开全文
• <div><p><a href="https://confluence.kiwi.com/display/MOB/Transportation">Docs</a></p>该提问来源于开源项目：kiwicom/mobile</p></div>
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...