hdu2444—The Accomodation of Students （二分图匹配）

最新推荐文章于 2021-09-14 21:40:10 发布

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本文探讨了如何通过二分图匹配算法解决学生宿舍分配问题，确保互不认识的学生被安排在同一宿舍，同时最大化室友间的相容性。文章首先介绍了如何判断给定的学生关系网络是否能被划分为两个互不相识的群体，接着使用二分匹配算法来计算最多可以安排多少对室友。此问题在社交网络分析、团队构建等领域有着广泛的应用。

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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

Sample Output

No
3

思路：先搜索判断是否为二分图，然后用二分匹配计算答案。


#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
int pan[210],mpp[210][210],mm[210][210],u[210];
int n,m,x,y,flag;
void dfs(int v,int p)
{
    for(int i=1; i<=n; i++)
    {
        if(mm[v][i])
        {
//            printf("%d %d &%d %d\n",v,i,p,pan[i]);
            mm[v][i]=mm[i][v]=0;
            if(pan[i]>=0&&pan[i]!=p%2)
            {
                flag=1;
            }
            pan[i]=p%2;
            dfs(i,p+1);
        }
    }
}
int found(int v)
{
    for(int i=1; i<=n; i++)
    {
        if(mpp[v][i]&&!u[i])
        {
            u[i]=1;
            if(!pan[i]||found(pan[i]))
            {
                pan[i]=v;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(pan,-1,sizeof(pan));
        memset(mpp,0,sizeof(mpp));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&x,&y);
            mpp[x][y]=1;
            mpp[y][x]=1;
        }
        memcpy(mm,mpp,sizeof(mm));
        flag=0;
        pan[1]=1;
        dfs(1,2);
        int ans=0;
        memset(pan,0,sizeof(pan));
        for(int i=1; i<=n; i++)
        {
            memset(u,0,sizeof(u));
            if(found(i))ans++;
        }
        if(flag)
            printf("No\n");
        else
            printf("%d\n",ans/2);
    }
}