hdu2444

B - The Accomodation of Students

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 41 21 31 42 36 51 21 31 42 53 6 

 

Sample Output

No
3 
题意：给一些人的关系，一，把他们放到两个屋子，每个屋子里的人彼此都不认识，能不能做到？，不就是二分图判断吗，染色法解决，二，如果能做到，是二分图，把他们两两分组，每组里的两个人必须认识，问最多分几组，不就是二分图最大匹配吗！
所以这道题就是问：我给你的图是二分图吗，是输出是，并且输出最大匹配数，不是就输出No,完事；
ac代码：
#include <iostream>#include <stdio.h>#include <vector>#include <string.h>#include <queue>using namespace std;const int maxn=1000;vector<int>G[maxn];int col[maxn];bool bfs(int s){    col[s]=1;    queue<int>que;    que.push(s);    while(!que.empty())    {        int u=que.front();        que.pop();        for(int i=0;i<G[u].size();i++)        {            int v=G[u][i];            if(col[v]==-1)            {                que.push(v);                col[v]=!col[u];            }            if(col[v]==col[u])                return false;        }    }    return true;}int used[maxn],linker[maxn];bool dfs(int u){    for(int i=0;i<G[u].size();i++)    {        int v=G[u][i];        if(!used[v])        {            used[v]=1;            if(linker[v]==-1||dfs(linker[v]))            {                linker[v]=u;                return true;            }        }    }    return false;}int solve(int n){    int res=0;    memset(linker,-1,sizeof(linker));    for(int u=1;u<=n;u++)    {        memset(used,0,sizeof(used));        if(dfs(u))res++;    }    return res/2;///加的双向边}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=-1)    {        int u,v;        for(int i=0;i<m;i++)        {            scanf("%d%d",&u,&v);            G[u].push_back(v);            G[v].push_back(u);        }        memset(col,-1,sizeof(col));        for(int i=1;i<=n;i++)        {            if(col[i]==-1&&!bfs(i))            {                cout<<"No"<<endl;                goto endW;            }        }        cout<<solve(n)<<endl;        endW:;        for(int i=1;i<=n;i++)            G[i].clear();    }    return 0;}