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

 

题意：有一些学生，他们之间有些互相认识，有些不认识，其中若A和B认识，B和C认识，并不代表A和C认识，现在提供互相认识的学生编号，问是否能将这些学生分成两组，同一组的学生互相不认识，如果可以，则分配给这些学生双人间，只有是朋友才能住双人间，问最多要准备几间房（

即问是否为二分图，若是，其最大匹配数是多少）

#include
         
          #include
          
           #include
           
            #include
            
             #define N 210using namespace std;int G[N][N], visit[N];int vis[N], use[N], col[N];int n, flag;void BFS(int u) //判断是否为二分图{    int i, v;    queue
             
              Q;    Q.push(u);    visit[u] = 1;    while (!Q.empty())    {        v = Q.front();        Q.pop();        for (i = 1; i <= n; i++)        {            if (v == i) continue;            if (!visit[i] && col[i] == 0 && G[v][i])            {                col[i] = -1;                visit[i] = 1;     //第一组的学生标记为1，他们的朋友标记为-1，若在查找中发现自己和朋友的标记一致，则说明不能分为互不认识的两组                Q.push(i);            }            else if (col[i] == col[v] && G[v][i])                flag = 1; //自己和朋友标记相同，则不是二分图        }    }}int Find(int u) //匈牙利算法查找最大匹配数{    int i;    for (i = 1; i <= n; i++)    {        if (!vis[i] && G[u][i])        {            vis[i] = 1;            if (!use[i] || Find(use[i]))            {                use[i] = u;                return 1;            }        }    }    return 0;}int main (){    int m, ans, i, a, b;    while (scanf("%d%d", &n, &m) != EOF)    {        memset(G, 0, sizeof(G));        memset(use, 0, sizeof(use));        memset(visit, 0, sizeof(visit));        memset(col, 0, sizeof(col));        ans = 0;        while (m--)        {            scanf("%d%d", &a, &b);            G[a][b] = 1;        }        flag = 0;        col[1] = 1;        BFS(1);        if (flag) printf("No\n");        else        {            for (i = 1; i <= n; i++)            {                memset(vis, 0, sizeof(vis));                if (Find(i)) ans++;            }            printf("%d\n", ans);        }    }    return 0;}