BZOJ 1433 假期的宿舍 二分图匹配

    这道题目不难，二分图匹配建模比较明显。加油吧！相信自己。（自己写的，好开心，40毫秒，比ccz略快）。 尽管算法模版是抄一本书上的，但这次很明显我是背出来的。不算抄。

#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#define rep(i,j,k) for(int i = j; i <= k; i++)
#define maxn 60
#define clr(i,j) memset(i,j,sizeof(i))
using namespace std;
 
vector<int> g[maxn];
int mx[maxn], my[maxn], dx[maxn], dy[maxn], n1;
queue<int> q;
bool vis[maxn];
 
bool match(int now)
{
    int s = g[now].size();
    rep(i,0,s-1){
        int to = g[now][i];
        if( !vis[to] && dy[to] == dx[now] + 1 ){
            vis[to] = 1;
            if( !my[to] || match(my[to]) ){
                my[to] = now, mx[now] = to;
                return 1;
            }
        }
    }
    return 0;
}
 
int erfen()
{
    clr(mx,0); clr(my,0); int ans = 0;
    while( 1 ){
        clr(dx,0), clr(dy,0);
        while( !q.empty() ) q.pop();
        rep(i,1,n1){
            if( !mx[i] ) q.push(i);
        }
        bool flag = 1;
        while( !q.empty() ){
            int now = q.front(); q.pop();
            int s = g[now].size();
            rep(i,0,s-1){
                int to = g[now][i];
                if( !dy[to] ){
                    dy[to] = dx[now] + 1;
                    if( my[to] ){
                        dx[my[to]] = dy[to] + 1;
                        q.push(my[to]);
                    }
                    else flag = 0;
                }
            }
        }
        if( flag ) break;
        clr(vis,0);
        rep(i,1,n1){
            if( !mx[i] && match(i) ) ans++;
        }
    }
    return ans;
}
 
void clear()
{
    rep(i,1,maxn) g[i].clear();
}
 
int read()
{
    int s = 0,t = 1; char c = getchar();
    while( !isdigit(c) ){
        if( c == '-' ) t = -1; c = getchar(); 
    }
    while( isdigit(c) ){
        s = s * 10 + c- '0'; c = getchar();
    }
    return s * t;
}
 
bool bed[maxn], home[maxn];
bool used[maxn][maxn];
 
int main()
{
    int t = read();
    while( t-- ){
        clear();
        clr(used,0);  clr(bed,0); clr(home,0);
        int n = read();
        n1 = n;
        rep(i,1,n){
            bed[i] = read();
        }
        int tot = 0;
        rep(i,1,n) {
            int x = read();
            if( bed[i] && x == 1 ) {
                home[i] = 1;  
                tot++;
            }
        }
        rep(i,1,n){
            rep(j,1,n){
                int x = read();
                if( i == j && bed[i] && !home[i] && !used[i][i] ) {
                     g[i].push_back(i);
                     used[i][i] = 1;
                }
                else if( i != j && x ){
                    if( bed[i] && !home[j] && !used[j][i] ) {
                        g[j].push_back(i);
                        used[j][i] = 1;
                    }
                    if( bed[j] && !home[i] && !used[i][j]) {
                        used[i][j] = 1;
                        g[i].push_back(j);
                    }
                }
            }
        }
        if( erfen() == n - tot ) printf("%c%c%c
", 94,95,94);
        else printf("%c%c%c
", 84,95,84);
    }
    
    return 0;
}