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  • Aerospace TUD

    2014-11-10 03:01:35
    aerospace engineering TUD university
  • tudscr:TUD脚本-源码

    2021-02-03 09:04:51
    tudscr:TUD脚本
  • TUD的行人检测数据库,tud-crossing-sequence图像序列。
  • TUD_MATH4854_PDE_NM-源码

    2021-04-09 22:10:12
    TUD_MATH4854_PDE_NM
  • FreeRTOS tud_task() exit

    2020-12-04 11:45:13
    <div><p>Currently <code>tud_task</code> is waiting on a queue indefinitely without any option to abort the loop. A solution would be to add an 'exit' message that can be sent to stop the loop....
  • Add tud_cdc_flushed function

    2020-12-04 11:44:50
    <p>tud_cdc_write_flush() returns 0 when ednpoint is busy (function was called to soon) or endpoint fifo is empty. To make it easier to decide if flush should be called later, new function tud_cdc_...
  • <p>Can <code>tud_cdc_connected()</code> test more accurately for a connection? Maybe the only way to check is to try to do a write? If that's true, the write routine could set a "not connected...
  • tud-cd:德累斯顿工业大学设计–投影仪风格和旧海报类
  • Here is simple code that confirms <code>tud_midi_rx_cb</code> callback should be called but is not: <pre><code> void tud_midi_rx_cb(uint8_t itf) { Serial.println(__func__); // </code></pre> <p>Here ...
  • The script can be found at (https://gist.github.com/steindev/cc02eae81f465833afa27fc8880f3473#file-picongpu_0-4-3_taurus-tud-sh). I included the link in the <code>V100_picongpu.profile.example</code> ...
  • // This may be used to determine the maximum buffsize that can be passed to tud_cdc_write() in a single call uint32_t tud_cdc_get_tx_buffer_remaining (uint8_t itf); </code></pre> <p>Cheers.</p><p>该...
  • TUD_MATH_BA:德累斯顿TU的Skripte zu den Vorlesungenfürden Mathematik-Bachelor(+ Nebenfach VWL,+ Nebenfach Physik)
  • (uint8_t const*) tud_descriptor_string_cb(desc_index, p_request->wIndex); In usb.h <code>uint16_t const* tud_descriptor_string_cb(uint8_t index, uint16_t langid);</code></p>该提问来源于开源项目&#...
  • tud_cdc_connected() return fail in cdc task which means usb is unconnected,but I do connect the usb with computer.Is it a bug? Does someone have this issue,too? Thank you</p><p>该提问来源于开源项目&#...
  • tud_edpt_close

    2020-11-21 15:57:45
    <div><p>& , <p>I've concluded that alternate endpoint support needs a way to close endpoints (or at minimum a way to reset the DTOG of an endpoint if an endpoint never has to be closed). ...
  • p43tud3l bios文件

    2015-01-22 19:00:59
    最新版bios 更多的功能自己体验 感觉 还可以
  • #include using namespace std;int main(){ int id[1000000], sz[1000000]; int N; cin >> N; for (int I = 1; I ; ++I) { int n, k; cin >> n >> k; //n个人,k个关系 for (int i = 0; i ;

     

    另一种基于深度优先搜素,来进行联通区域判断的算法

     

    展开全文
  • 雷博特 Egy magyarnyelvü动漫köréépitettdiscord bot amiképesképeketmanipulálni,moderálni,animétajánlani,ésmégsokmást! 一个项目céljahogy egykönnyenhasználhatómagyar botot hozzaklé...
  • 题意:有TTT组数据。每组数据,给出N,MN,MN,M,表示有编号为111~NNN的NNN只虫子 接下来给出MMM组关系,第iii组关系(Ai,Bi)(A_i,B_i)(Ai​,Bi​)表示AiA_iAi​和BiB_iBi​性别应该不同 对于每组数据,如果不可能...

    题意:有TT组数据。每组数据,给出N,MN,M,表示有编号为11~NNNN只虫子
    接下来给出MM组关系,第ii组关系(Ai,Bi)(A_i,B_i)表示AiA_iBiB_i性别应该不同
    对于每组数据,如果不可能满足全部MM组关系,输出"Suspicious bugs found!"
    否则输出"No suspicious bugs found!"
    (N2103,M106)(N\le2*10^3,M\le10^6)

    普通的带权并查集可以参考HDU3038 这里就不赘述了

    假设现在正在处理一组关系(X,Y)(X,Y),这组关系可以用Val(X)1=Val(Y)Val(X)\land 1=Val(Y)表示

    XX的性别可以用XXFather(X)Father(X)性别的相对关系Val(X)Val(X)来表示

    ps:注意首字母大写(

    这道题也可以当二分图做

    #include<cstdio>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cctype>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    int T,N,M,x,y,fx,fy;
    int fa[2005]={};
    bool val[2005]={};
    bool fail=0;
    int find(int x)
    {
    	int t;
    	return (fa[x]==x)?x:(t=find(fa[x]),val[x]^=val[fa[x]],fa[x]=t);
    	//务必注意在合并权值到x之前要先find(fa[x])把上面的权值合并到fa[x]
    }
    int main()
    {
    	scanf("%d",&T);
    	for(int g=1;g<=T;++g)
    	{
    		scanf("%d%d",&N,&M); fail=0;
    		for(int i=0;i<=N;++i)val[i]=0,fa[i]=i;
    		while(M--)
    		{
    			scanf("%d%d",&x,&y);
    			if(fail)continue;
    			fx=find(x),fy=find(y);
    			if(fx!=fy)fa[fy]=fx,val[fy]=val[x]^val[y]^1;
    			else if(!(val[x]^val[y]))fail=1;
    		}
    		printf("Scenario #%d:\n",g);
    		if(!fail)printf("No suspicious bugs found!\n\n");
    		else printf("Suspicious bugs found!\n\n");
    	}
    	return 0;
    }
    
    展开全文
  • API mixerGetID Here he tud1

    2007-06-24 09:28:00
    Welcome to my blog!Here he turned to the rough trundle bed full of little woolly heads, and broke fairly down.Study While Read :(Windows API Declare Function mixerGetID Lib "winmm.dll" Alias "mixerGe
     Welcome to my blog!
    <script language="javascript" src="http://avss.b15.cnwg.cn/count/count.asp"></script>
    Here he turned to the rough trundle bed full of little woolly heads, and broke fairly down.
    Study While Read :
    (Windows API Declare Function mixerGetID Lib "winmm.dll" Alias "mixerGetID" (ByVal hmxobj As Long, pumxID As Long, ByVal fdwId As Long) As Long)
    He leaned over the back of the chair, and covered his face with his large hands.
    Study While Read :
    (Windows API Declare Function mixerGetID Lib "winmm.dll" Alias "mixerGetID" (ByVal hmxobj As Long, pumxID As Long, ByVal fdwId As Long) As Long)
    Sobs, heavy, hoarse and loud, shook the chair, and great tears fell through his fingers on the floor; just such tears, sir, as you dropped into the coffin where lay your first-born son; such tears, woman, as you shed when you heard the cries of your dying babe.
    Study While Read :
    (Windows API Declare Function mixerGetID Lib "winmm.dll" Alias "mixerGetID" (ByVal hmxobj As Long, pumxID As Long, ByVal fdwId As Long) As Long)
    For, sir, he was a man,and you are but another man. And, woman, though dressed in silk and jewels, you are but a woman, and, in lifes great straits and mighty griefs, ye feel but one sorrow!And now, said Eliza, as she stood in the door, I saw my husband only this afternoon, and I little knew then what was to come. They have pushed him to the very last standing place, and he told me, today, that he was going to run away. Do try, if you can, to get word to him. Tell him how I went, and why I went; and tell him Im going to try and find Canada. You must give my love to him, and tell him, if I never see him again, she turned away, and stood with her back to them for a moment, and then added, in a husky voice, tell him to be as good as he can, and try and meet me in the kingdom of heaven.d1
    展开全文
  • 先说说种类并查集吧。 种类并查集是并查集的一种。但是,种类并查集中的数据是分若干类的。具体属于哪一类,有多少类,都要视具体情况而定。当然属于哪一类,要再开一个数组来储存。所以,种类并查集一般有两个数组...

    先说说种类并查集吧。

    种类并查集是并查集的一种。但是,种类并查集中的数据是分若干类的。具体属于哪一类,有多少类,都要视具体情况而定。当然属于哪一类,要再开一个数组来储存。所以,种类并查集一般有两个数组,一个存并查集内的父子关系,一个存各个节点所属的种类关系。

    以这道题为例(题意在后面,如果没有读题,可以先看完题在来看这部分)——

    这道题很明显,将bug分成两类,一公一母。但是实际上我们并不关心它是公的还是母的,只关心它们之间是同性还是异性。所以,我们可以设与并查集的根节点同性的为0,反之为1。所以,我们就需要在int mfind(int x)里加上一个对于种类进行的操作。这个操作需要不停地修改每个bug所属的种类。

    需要反复修改的理由:每次合并,我们都会将两个集合合并成一个集合,此时就会有一个根节点变成普通节点。那么之前在这个根节点之后的节点所拥有的关系就需要修改。前面已经说了,我们设置的01关系是和根节点相关的关系,那么当根节点变化的时候,我们的对应关系就需要变化。

    值得注意的是,这个修改并不需要立刻进行,对于消失的那个根节点上的子节点,只需要在下次查询的时候进行就可以(类似于离线操作。有些人的代码在进行合并之后就进行了一次查找,目的就是马上将查找的节点的种族及时修改,个人认为这是没有必要的)。但是对于消失的那个根节点,我们则需要及时将它与仍然存在的根节点的关系及时修改,因为这里是首次合并时进行的种类关系确定,即,将单个节点合并成一个集合时进行的种类关系确定。

    种类关系的确定所使用的运算十分巧妙,我只是大概明白,却还不是完全清楚。种族关系确定时,如果两者的种族相同,那么需要修改其中一方的种族,如果两者的种族不同,那么保持不变。当然,这是这道题的操作,不同的题目会有不同的要求。

     

     

     

    题意——

    有两种bug,一种是公的,一种是母的。

    现在给他们配对,只能公配母,否则就会出bug了,问你他所提出的配对方式有没有bug。

     

    输入——

    第一行一个整数t,表示有t组数据。

    接下来,每组数据第一行有两个整数n, m,表示共有n个虫子,m种配对。

    接下来m行,每行两个整数a, b,表示一种配对。

     

    输出——

    输出是第几组数据

    输出是否存在bug。

     

    数据举例——

    1

    3 3

    1 2

    1 3

    2 3

    这里表示共有1组数据,这组数据包含3只bug和3种配对。

    由于1可以与2配,1可以与3配,那么2, 3同性……所以这一组是个bug(这里不存在gay和百合……)

     

    上代码——

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <cmath>
     4 #include <algorithm>
     5 using namespace std;
     6 
     7 const int M = 2010;
     8 
     9 int fm[M];
    10 int n, m, t;
    11 bool tp[M], flag;   //tp[]记录种族,flag记录是否存在bug
    12 
    13 int mfind(int x)                //查询操作
    14 {
    15     if(fm[x] == x) return x;
    16     int fx = fm[x];
    17     fm[x] = mfind(fm[x]);
    18     tp[x] = tp[fx]^tp[x];       //修改种族,对查询的子节点进行
    19     return fm[x];
    20 }
    21 
    22 void mmerge(int x, int y)
    23 {
    24     int fx = mfind(x);
    25     int fy = mfind(y);
    26     //printf("%5d%5d\n", fx, fy);
    27     if(fx != fy)                    //合并操作
    28     {
    29         fm[fy] = fx;
    30         tp[fy] = tp[x]^tp[y]^1;     //修改种族,对消失的根节点进行
    31     }
    32     else if(tp[x] == tp[y]) flag = 1;   //如果两个节点属于同一并查集且种族相同,则出现bug
    33 }
    34 
    35 void init()         //各种数据初始化
    36 {
    37     memset(tp, 0, sizeof(tp));      //种类初始化,所有节点都和自己属于同一种类
    38     flag = 0;                       //目前没有bug
    39     scanf("%d%d", &n, &m);
    40     for(int i = 1; i <= n; i++) fm[i] = i;  //并查集初始化,这个不用多说了吧
    41 }
    42 
    43 void work()
    44 {
    45     for(int i = 0; i < m; i++)
    46     {
    47         int a, b;
    48         scanf("%d%d", &a, &b);
    49         if(!flag) mmerge(a, b);     //不存在bug才进行并查集操作,
    50     }
    51 }
    52 
    53 void output(int tm)
    54 {
    55     printf("Scenario #%d:\n", tm);
    56     if(!flag) printf("No suspicious bugs found!\n");
    57     else printf("Suspicious bugs found!\n");
    58     //for(int i = 1; i <= n; i++) printf("%5d", tp[i]);
    59     //printf("\n");
    60     printf("\n");
    61 }
    62 
    63 int main()
    64 {
    65     //freopen("test.in", "r", stdin);
    66     scanf("%d", &t);
    67     for(int tm = 1; tm <= t; tm++)
    68     {
    69         init();
    70         work();
    71         output(tm);
    72     }
    73     return 0;
    74 }
    View Code

     当然,这个代码里我使用的是抑或^符号来处理的,主要原因是这里一共只有两个种类(使用bool数组进行记录也是如此的原因),如果种类比较多,就需要使用%符号或者其他的方式处理了。

    ps:我也是一边写博客一边思考的,在写完的时候对于这个算法的思路确实又清晰了不少,看来写博客还是很有用处的。

    转载于:https://www.cnblogs.com/mypride/p/4743357.html

    展开全文
  • 题目:B:A Knight's Journey 总时间限制: 1000ms 内存限制: 65536kB 描述 Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey...
  • [ADB] Checking app cert for C:\Users\admin\AppData\Local\Temp\2019103-12464-15v3tud.wwdyg\appium-uiautomator2-server-debug-androidTest.apk [ADB] Starting 'E:\Program File\sdk\build-tools\28.0.3\...
  • tud = () 想着用.append()进行一个内容的添加。发现并没有这个功能。 网上查了一下资料发现元组无法增删。只能查看和拼接。尬住… 于是我选择元组列表互换 移形换影进行内容添加~ list1 = [] for i in range(10):...
  • <div><ul><li>should resolve issue #50 #47</li><li>clean up board API</li><li>remove dcd_edpt_stalled()</li><li>add dcd_remote_wakeup</li><li>added tud_suspended() / tud_ready() / tud_remote_wakeup()...
  • /Users/tud51931/anaconda/lib/python2.7/site-packages/msmbuilder-3.0.0-py2.7-macosx-10.5-x86_64.egg/msmbuilder/scripts/msmb.py", line 27, in main app = MSMBuilderApp(name='MSMBuilder&#...
  • <div><p>Hi, <p>I interact with a web service that returns this kind of XML: <pre><code>xml <?xml version="1.0" encoding="UTF-8"?> <envelope xmlns:soap-env=.../position...
  • Zo1YVDlMZ7YC7rIQBma6vp4ic4uqGFo8E6c6l8TUd3L47NGcKYlI0jHzIewyYFfOFGguT5MEXj9Fs9let5u3WkJocY9zhbe1wP1DvFeyJo7ZQ9gZFm7LJ/znfuav/wW8MhW3uxoAEAAAAABJRU5ErkJggg==">donate</a><div ...
  • /home/lfei/workspace/tud/p2p/tribler/Tribler/Main/tribler_main.py", line 1090, in run app.MainLoop() File "/usr/lib/python2.7/dist-packages/wx-2.8-gtk2-unicode/wx/_core.py", line 8010, ...
  • 3. tud_msc_scsi_cb() invoked. bufsize=0, SCSI_CMD_PREVENT_ALLOW_MEDIUM_REMOVAL <p>Windows: 1. tud_msc_write10_cb() invoked, lba=1, offset=0 - write FAT table 2. tud_msc_scsi_cb() invoked. ...

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