HDU 2444

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3003    Accepted Submission(s): 1407

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

2008 Asia Harbin Regional Contest Online

 

Recommend

gaojie

/**       
          题意：n个人中有m组数据，表示两个人彼此熟悉，然后看怎么分可以使得现在的
          组内的人是彼此互不了解的
          做法：二分图匈牙利算法
**/
#include<iostream>
#include<cmath>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<queue>>
using namespace std;
#define maxn 210
int g[maxn][maxn];
int linker[maxn];
bool  used[maxn];
int un,vn;
int n;
int color[maxn];
int  bfs()
{
          queue<int>que;
          memset(color,-1,sizeof(color));
          que.push(1);
          color[1] = 0;
          while(!que.empty())
          {
                    int tt = que.front();
                    que.pop();
                    int mm = 1 - color[tt];
                    for(int i=1;i<=n;i++)
                    {
                              if(g[tt][i])
                              {
                                        if(color[i] == -1)
                                        {
                                                  color[i] = mm;
                                                  que.push(i);
                                        }
                                        else
                                        {
                                                  if(color[i] == color[tt]) return 0;
                                        }
                              }
                    }
          }
          return 1;
}
bool dfs(int u)
{
    for(int v = 1; v <= vn; v++)
    {
        if(g[u][v] && used[v] == false)
        {
            used[v] =  true;
            if(linker[v] == -1 || dfs(linker[v]))
            {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int res = 0;
    memset(linker,-1,sizeof(linker));
    for(int i=1; i<=un; i++)
    {
        memset(used,false,sizeof(used));
        res += dfs(i);
    }
    return res;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int m;
    while(~scanf("%d %d",&n,&m))
    {
        memset(g,0,sizeof(g));

        int u,v;
        for(int i=0; i<m; i++)
        {
            scanf("%d %d",&u,&v);
            g[u][v] = 1;
            g[v][u] = 1;
        }
        un = vn = n;
        if(!bfs())  printf("No
");
        else printf("%d
",hungary()/2);
    }
    return 0;
}