The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5931 Accepted Submission(s): 2703

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  

题目大意:

给你一张图，判断是否能构成二分图，如果能求出最大匹配

题目思路：

可以直接bfs进行染色，如果在同一集合中的点的颜色不相同则说名不能构成，否则能够成二分图然后进行最大匹配

AC代码：

#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
const int maxn = 250;
using std::vector;
using std::queue;
vector<int>edge[maxn];

bool vis[maxn];
int link[maxn];
int n,m;

bool dfs(int u){
    for(int i=0;i<edge[u].size();i++){
        int v = edge[u][i];
        if(!vis[v]){
            vis[v]=true;
            if(link[v]==-1||dfs(link[v])){
                link[v]=u;
                return true;
            }
        }
    }
    return false;
}

bool bfs(int u){           //bfs染色判断
    link[u]=0;
    queue<int>q;
    q.push(u);
    while(!q.empty()){
        u = q.front();q.pop();
        for(int i=0;i<edge[u].size();i++){
            int v = edge[u][i];
            if(link[v]==-1){link[v]=link[u]^1;q.push(v);}
            else {
                if(link[v]==link[u])return false;   //在不同集合中且颜色相同
            }
        }
    }
    return true;
}

int main()
{

    while(~scanf("%d%d",&n,&m)){
        memset(link,-1,sizeof(link));
        memset(edge,0,sizeof(edge));
        while(m--){
            int u,v;scanf("%d%d",&u,&v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
        int flag=0;
        for(int i=1;i<=n;i++){
            if(link[i]==-1){
                if(!bfs(i)){
                    flag=1;
                    break;
                }
            }
        }
        if(flag){
            printf("No\n");
            continue;
        }
        memset(link,-1,sizeof(link));
        int ans = 0;
        for(int i=1;i<=n;i++){
            memset(vis,false,sizeof(vis));
            if(dfs(i))ans++;
        }
        printf("%d\n",ans/2);
    }
    return 0;
}

HDU2444-判断是否能构成二分图

The Accomodation of Students