HDU3081 - Marriage Match II - 最大匹配+并查集+匈牙利算法

AC代码

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
#define inf 0x3f3f3f3f

const int N=300;
int n,match[N],f[N];
bool book[N],e[N][N];

int getf(int x)
{
    if(f[x]==x)
        return x;
    return f[x]=getf(f[x]);
}

void merge(int x,int y)
{
    int t1=getf(x);
    int t2=getf(y);
    f[t2]=t1;
}

bool dfs(int x)
{
    for(int i=1; i<=n; i++)
    {
        if(book[i]==0&&e[x][i])
        {
            book[i]=1;
            if(match[i]==-1||dfs(match[i]))
            {
                match[i]=x;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int m,F,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d",&n,&m,&F);
        for(int i=1; i<=n; i++)
            f[i]=i;
        memset(e,0,sizeof(e));//注意清空，不然WA
        memset(match,-1,sizeof(match));//清空。。
        for(int i=1; i<=m; i++) //girl and boy never quarrel
        {
            int x,y;
            scanf("%d %d",&x,&y);
            e[x][y]=1;
        }
        for(int i=1; i<=F; i++)//x and y are friends
        {
            int x,y;
            scanf("%d %d",&x,&y);
            merge(x,y);
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                int t1=getf(f[i]);
                int t2=getf(f[j]);
                if(t1==t2)
                {
                    for(int k=1; k<=n; k++)
                    {
                        if(e[i][k]||e[j][k])
                            e[i][k]=e[j][k]=1;
                    }
                }
            }
        }
        int ans=0;
        while(1)
        {
            memset(match,-1,sizeof(match));
            int w=0;
            for(int i=1; i<=n; i++)
            {
                memset(book,0,sizeof(book));
                w+=dfs(i);
            }
            if(w==n)
            {
                ans++;
                for(int i=1; i<=n; i++)
                {
//                    if(match[i]!=-1)
                        e[match[i]][i]=0;
                }
            }
            else
                break;
        }
        printf("%d\n",ans);
    }
    return 0;
}

参考博客

https://blog.csdn.net/jingdianitnan/article/details/24151631?depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-7&utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-7