POJ 2594 Treasure Exploration（最小可相交路径覆盖）题解

题意：有n个点，m条单向边，每个机器人能沿着单向边走，能重复经过一个点，问最少几个机器人走遍n个点

思路：原来以前学的都是不能相交的算法....可相交的做法是跑Floyd把能到达的都加上边，然后跑最小覆盖

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 500 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int linker[maxn], n, m;
int g[maxn][maxn];
bool used[maxn];
bool dfs(int u){
    for(int v = 1; v <= n; v++){
        if(g[u][v] && !used[v]){
            used[v] = true;
            if(linker[v] == -1 || dfs(linker[v])){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int hungry(){
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for(int u = 1; u <= n; u++){
        memset(used, false, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}
void floyd(){
    for(int k = 1; k <= n; k++){
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(g[i][k] && g[k][j])
                    g[i][j] = 1;
            }
        }
    }
}
int main(){
    while(~scanf("%d%d", &n, &m) && n + m){
        memset(g, 0, sizeof(g));
        while(m--){
            int u, v;
            scanf("%d%d", &u, &v);
            g[u][v] = 1;
        }
        floyd();
        printf("%d
", n - hungry());
    }
    return 0;
}