【二分图最大匹配】Bullet @山东省第九届省赛 B

时间限制: 6 Sec 内存限制: 128 MB
题目描述
In GGO, a world dominated by gun and steel, players are fighting for the honor of being the strongest gunmen. Player Shino is a sniper, and her aimed shot kills one monster at a time. Now she is in an n × n map, and there are monsters in some grids. Each monster has an experience. As a master, however, Shino has a strange self-restrain. She would kill at most one monster in a column, and also at most one in a row. Now she wants to know how to get max experience, under the premise of killing as many monsters as possible.
输入
The first line contains an integer n
Then n lines follow. In each line there are n integers, and Aij represents the experience of the monster at grid(i,j). If Aij=0, there is no monster at grid(i,j).
输出
One integer, the value of max experience.
样例输入
2
2 0
1 8
样例输出
2

每行每列最多取一个数，构成的集合在满足元素数量最多的前提下，最小值最大。
最小值增大时，可匹配边的数量减少，所以最大匹配可能减小，于是可以二分最小值，每次求图中权值大于最小值的边的最大匹配。

#define IN_LB() freopen("C:\\Users\\acm2018\\Desktop\\in.txt","r",stdin)
#define OUT_LB() freopen("C:\\Users\\acm2018\\Desktop\\out.txt","w",stdout)
#define IN_PC() freopen("C:\\Users\\hz\\Desktop\\in.txt","r",stdin)
#include <bits/stdc++.h>
using namespace std;

const int maxn = 505;
struct edge {
    int v,w,nex;
} ed[maxn*maxn];
int head[maxn],cnt,n,vis[maxn],match[maxn];
void addedge(int u,int v,int w) {
    cnt++;
    ed[cnt].v = v;
    ed[cnt].w = w;
    ed[cnt].nex = head[u];
    head[u] = cnt;
}

bool dfs(int u,int limit) {
    for(int i=head[u]; i; i=ed[i].nex) {
        int v = ed[i].v;
        if(ed[i].w>=limit&&!vis[v]) {
            vis[v] = 1;
            if(!match[v]||dfs(match[v],limit)) {
                match[v] = u;
                return true;
            }
        }
    }
    return false;
}

int judge(int limit) {
    memset(match,0,sizeof match);
    int cnt = 0;
    for(int i=1; i<=n; i++) {
        memset(vis,0,sizeof vis);
        if(dfs(i,limit))
            cnt++;
    }
    return cnt;
}

int main() {
//    IN_LB();
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) {
            int weight;
            scanf("%d",&weight);
            addedge(i,j,weight);
        }
    }
    int ans = judge(1);
    int res = 0,base = 1<<30;
    while(base>=1) {
        if(judge(res+base) == ans) {
            res += base;
        } else
            base >>= 1;
    }
    printf("%d\n",max(1,res));
    return 0;
}