洛谷P2055假期的宿舍

题目

此题主要是考察二分图匹配，而二分图匹配最主要的就是建图，而图一般都是要分成两个部分来分，比如该题就需要先将在学校住的人和床连在一起，因为在学校住就会与一个床。然后每两个人之间假如他们相互认识就可以相互连边，这样就可以通过人人相连实现人床相连。

\(Code\)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
int n1, T, cnt;
int map[1010][1010];
int vis[200100], lin[200010], school[200100], home[200010], flag[200100];
struct cym {
    int to, nex;
}e[200010];
inline void add(int f, int t)
{
    e[++cnt].to = t;
    e[cnt].nex = lin[f];
    lin[f] = cnt;
}
inline bool dfs(int now)
{
    for (int i = lin[now]; i; i = e[i].nex)
        if (!vis[e[i].to])
        {
            vis[e[i].to] = 1;
            if (!flag[e[i].to] || dfs(flag[e[i].to])) {
                flag[e[i].to] = now; return true;
            }
        }
    return false;
}
int main()
{
    scanf("%d", &T);
    while (T--)
    {
        int ans = 0, sum = 0;
        memset(map, 0, sizeof(map)); memset(school, 0, sizeof(school));memset(home, 0, sizeof(home));memset(flag, 0, sizeof(flag));memset(e, 0, sizeof(e));memset(lin, 0, sizeof(lin));
        cnt = 0;
		scanf("%d", &n1);
        for (int i = 1; i <= n1; i++)
            scanf("%d", &school[i]);
        for (int i = 1; i <= n1; i++)
        {
            scanf("%d", &home[i]);
            if (!school[i])
            	home[i] = 0;
            if (school[i])
                add(i, i);
        }
        for (int i = 1; i <= n1; i++)
        	if (!home[i])
        		sum++;
        for (int i = 1; i <= n1; i++)
            for (int j = 1; j <= n1; j++)
            {
                scanf("%d", &map[i][j]);
                if (map[i][j] && school[j])
                    add(i, j);
            }
        for (int i = 1; i <= n1; i++)
            if (!home[i])
            {
                memset(vis, 0, sizeof(vis));
                if (dfs(i)) ans++;
            }
        if (ans == sum) printf("^_^\n");
        else printf("T_T\n");
    }
    return 0;
}
/*
2
3
1 1 0
0 1 0
0 0 1
0 0 0
1 0 0
3
1 1 0
0 1 0
0 1 1
1 0 0
1 0 0
*/