zoukankan      html  css  js  c++  java
  • HDU——T 2444 The Accomodation of Students

    http://acm.hdu.edu.cn/showproblem.php?pid=2444

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7087    Accepted Submission(s): 3168


    Problem Description
    There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

    Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

    Calculate the maximum number of pairs that can be arranged into these double rooms.
     
    Input
    For each data set:
    The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

    Proceed to the end of file.

     
    Output
    If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
     
    Sample Input
    4 4 1 2 1 3 1 4 2 3 6 5 1 2 1 3 1 4 2 5 3 6
     
    Sample Output
    No 3
     
    Source
     
    Recommend
    gaojie   |   We have carefully selected several similar problems for you:  2438 2443 2442 2441 2440 
     
     
    蒟蒻无脑法:二分图染色+二分图匹配模板
    代码好诡异
      1 #include <algorithm>
      2 #include <cstring>
      3 #include <cstdio>
      4 #include <queue>
      5 
      6 using namespace std;
      7 
      8 const int N(233);
      9 int n,m,head[N],sumedge;
     10 struct Edge
     11 {
     12     int v,next;
     13     Edge(int v=0,int next=0):v(v),next(next){}
     14 }edge[233333];
     15 inline void ins(int u,int v)
     16 {
     17     edge[++sumedge]=Edge(v,head[u]);
     18     head[u]=sumedge;
     19     edge[++sumedge]=Edge(u,head[v]);
     20     head[v]=sumedge;
     21 }
     22 
     23 int col[N];
     24 bool Paint(int x)
     25 {
     26     col[x]=0;
     27     queue<int>que;
     28     que.push(x);
     29     for(int u,v;!que.empty();)
     30     {
     31         u=que.front(); que.pop();
     32         for(int i=head[u];i;i=edge[i].next)
     33         {
     34             v=edge[i].v;
     35             if(col[v]!=-1)
     36             {
     37                 if(col[v]==col[u])
     38                     return false;
     39             }
     40             else
     41             {
     42                 col[v]=col[u]^1;
     43                 que.push(v);
     44             }
     45         }
     46     }
     47     return true;
     48 }
     49 
     50 int sumvis,vis[N],match[N],Map[N][N];
     51 bool find(int u)
     52 {
     53     /*for(int v,i=head[i];i;i=edge[i].next)
     54     {
     55         v=edge[i].v;
     56         if(!vis[v])
     57         {
     58             vis[v]=1;
     59             if(!match[v]||find(match[v]))
     60             {
     61                 match[v]=u;
     62                 return true;
     63             }
     64         }
     65     }*/
     66     for(int v=1;v<=n;v++)
     67         if(Map[u][v]&&!vis[v])
     68         {
     69             vis[v]=1;
     70             if(!match[v]||find(match[v]))
     71             {
     72                 match[v]=u;
     73                 return true;
     74             }
     75         }
     76     return false;
     77 }
     78 
     79 inline void init()
     80 {
     81     memset(edge,0,sizeof(edge));
     82     memset(head,0,sizeof(head));
     83     memset(match,0,sizeof(match));
     84 }
     85 
     86 int main()
     87 {
     88     for(;~scanf("%d%d",&n,&m);init())
     89     {
     90         int ans=0,flag=0;sumvis=0;
     91         memset(Map,0,sizeof(Map));
     92         for(int u,v;m--;ins(u,v))
     93             scanf("%d%d",&u,&v),Map[u][v]=1;
     94         memset(col,-1,sizeof(col));
     95         for(int i=1;i<=n;i++)
     96             if(col[i]==-1)
     97                 if(!Paint(i))
     98                 {
     99                     flag=1;
    100                     break;
    101                 }
    102         if(flag)
    103         {
    104             puts("No");
    105             continue;
    106         }
    107         for(int i=1;i<=n;i++)
    108         {
    109             if(find(i)) ans++;
    110             memset(vis,0,sizeof(vis));
    111         }
    112         printf("%d
    ",ans);
    113     }
    114     return 0;
    115 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
  • 相关阅读:
    Appium自动化环境搭建
    真机Android 8.0版本以上uiautomator定位元素-Unsupported protocol: 2/Unexpected error while obtaining UI hierarchy错误处理
    rsa非对称加密
    QT使用OpenSSL的接口实现RSA的加密解密
    lua安装后其他库使用产生问题解决方法
    log4cpp的使用描述
    std::function和std::bind
    C++11线程睡眠的方式
    高精度计时器
    如何解决TCP拆包粘包问题
  • 原文地址:https://www.cnblogs.com/Shy-key/p/7435403.html
Copyright © 2011-2022 走看看