二分图

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

判断该图是否是二分图 不是输出no 如果是输出最大匹配值

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<iostream>
using namespace std;
const int N=1e6+10;
int map[300][300];
int book[300] ;   //染色
int mat[300];  //匹配
int flag;
int n,m,sum;
int match(int x)   //匈牙利   找最大匹配值
{
    for(int i=1;i<=n;i++)
    {
        if(map[x][i]&&book[i]==0)  //如果连通 但没标记过
        {
            book[i]=1;
            if(!mat[i]||match(mat[i])) //如果没匹配过 或者能腾出个位置   如果1能和2匹配  而2已经和3匹配了  那我们就看3能不能和别人匹配 如果3能和别人匹配了 那么1就能和2匹配
            {
                mat[i]=x;
                return 1;
            }
        }
    }
      return 0;
}
void dfs(int x ,int clore)   ///染色法判读是否是二分图
{
     ///flag=0;
    book[x]=clore;
    if(flag==1)
        return ;
    for(int i=1;i<=n;i++)
    {
        if(map[x][i]==1)
        {
            if(book[i]==0)
            {
                dfs(i,-clore);
            }
             else if(book[x]==book[i])
             {
                 flag=1;
                 return ;
             }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        sum=0;flag=0;
        memset(book,0,sizeof(book));
        memset(mat,0,sizeof(mat));
        memset(map,0,sizeof(map));
       for(int i=0;i<m;i++)
       {
           int a,b;
           scanf("%d %d",&a,&b);
           map[a][b]=1;
           map[b][a]=1;
       }
       dfs(1,1);   //  从第一个点开始  染成1这个颜色
       if(flag==1)
        printf("No\n");
       else
       {
           for(int i=1;i<=n;i++)
           {
               memset(book,0,sizeof(book));   //每一步都要清空   每一个人都可以和另一组的人重新匹配
                if(match(i))
                    sum++;
                //printf("$$$$$%d\n",sum);
           }
           printf("%d\n",sum/2);
       }
    }
}