poj 2594 Treasure Exploration (floyd传递闭包+最小路径覆盖) (bitset优化floyd)

这道题为有向图有相交边的情况。。不能直接求最大匹配

先用floyd处理一下边

//
//  main.cpp
//  wzazzy
//
//  Created by apple on 2018/10/23.
//  Copyright © 2018年 apple. All rights reserved.
//

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long int ll;
const int maxn =1500+10;
const int maxm=10000;
const int mod =1e9+7;
const int INF=0x3f3f3f3f;
//**********************************
int n, m, maze[510][510], linker[510];
bool used[510];

void Floyd(){
    for(int k = 1; k <= n; ++k)
        for(int i = 1; i <= n; ++i)
            if(maze[i][k]){
                for(int j = 1; j <= n; ++j)
                    if(maze[k][j]) maze[i][j] = 1;
            }
}

bool dfs(int u){
    for(int v = 1; v <= n; ++v){
        if(maze[u][v] && !used[v]){
            used[v] = true;
            if(linker[v] == -1 || dfs(linker[v])){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungary(){
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for(int i = 1; i <= n; ++i){
        memset(used, false, sizeof(used));
        if(dfs(i)) ++res;
    }
    return res;
}

int main()
{
    while(~scanf("%d%d", &n, &m) && (n||m)){
        memset(maze, 0, sizeof(maze));
        int u, v;
        for(int i = 0; i < m; ++i){
            scanf("%d%d", &u, &v);
            maze[u][v] = 1;
        }
        Floyd();
        int ans = hungary();
        printf("%d\n", n-ans);
    }
}

还可以更快，使用bitset优化。代码如下

//
//  main.cpp
//  wzazzy
//
//  Created by apple on 2018/10/23.
//  Copyright © 2018年 apple. All rights reserved.
//

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<bitset>
#include<vector>
using namespace std;
typedef long long int ll;
const int maxn =1500+10;
const int maxm=10000;
const int mod =1e9+7;
const int INF=0x3f3f3f3f;
//**********************************
int n, m, linker[510];
bitset<510> maze[510];
bool used[510];
void floyd()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(maze[j][i]) maze[j]|=maze[i];
}
bool dfs(int u){
    for(int v = 1; v <= n; ++v){
        if(maze[u][v] && !used[v]){
            used[v] = true;
            if(linker[v] == -1 || dfs(linker[v])){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungary(){
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for(int i = 1; i <= n; ++i){
        memset(used, false, sizeof(used));
        if(dfs(i)) ++res;
    }
    return res;
}

int main()
{
    while(~scanf("%d%d", &n, &m) && (n||m)){
        memset(maze, 0, sizeof(maze));
        int u, v;
        for(int i = 0; i < m; ++i){
            scanf("%d%d", &u, &v);
            maze[u][v] = 1;
        }
        floyd();
        int ans = hungary();
        printf("%d\n", n-ans);
    }
}