B

B - The Accomodation of Students
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

/*此题主要分为两部分内容，第一部分是把所有的数据分为二分图，如果无法分成二分图就打印No并换数据，如果成功就计算最大匹配并输出*/#include <iostream>#include <vector>#include <cstring>#include <cstdio>#include <queue>using namespace std;const int N = 200;int n,m;vector<int > G[N];int used[N],match[N],Mat[N];void add_edge(int a,int b){    G[a].push_back(b);    G[b].push_back(a);return;}//第一部分 匹配 把所有的数据分为二分图//并用match[]用0 1分别表示不同的组int bine(){    memset(match,-1,sizeof(match));    for(int i=1;i<=n;i++)    {        if(match[i]<0){            queue<int> que;            que.push(i);            while(!que.empty())            {                int q=que.front();que.pop();                if(match[q]<0){                    match[q]=0;                }                for(int j=0;j<G[q].size();j++)                {                    int &e=G[q][j];                    if(match[e]<0){                        match[e]=match[q]^1;                        que.push(e);                    }                    else if(match[e]!=match[q]){                        match[e]=match[q]^1;                    }else                        return 1;                }            }        }    }    return 0;}//第二部分计算最大匹配bool dfs(int a){    used[a]=1;    for(int i=0;i<G[a].size();i++)    {        int &e=G[a][i];int w=Mat[e];        if(w<0||!used[w]&&dfs(w)){            Mat[a]=e;            Mat[e]=a;            return true;        }    }    return false;}int bipartite(){    int f=0;memset(Mat,-1,sizeof(Mat));    for(int i=1;i<=n;i++)    {        if(match[i]==0)        {            memset(used,0,sizeof(used));            if(dfs(i))                f++;        }    }    return f;}int main(){    while(~scanf("%d%d",&n,&m))    {        int a,b;        for(int i=0;i<=n;i++)            G[i].clear();        for(int i=0;i<m;i++)        {            cin>>a>>b;            add_edge(a,b);        }        if(bine()){            printf("No\n");            continue;        }        int ans=bipartite();        cout<<ans<<endl;    }    return 0;}