POJ 2594--Treasure Exploration【二分图 && 最小路径覆盖 && 点可以重复走 && 传递闭包】

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7343		Accepted: 3002

Description

Have you ever read any book about treasure exploration? Have you ever see any film about treasure exploration? Have you ever explored treasure? If you never have such experiences, you would never know what fun treasure exploring brings to you. 
Recently, a company named EUC (Exploring the Unknown Company) plan to explore an unknown place on Mars, which is considered full of treasure. For fast development of technology and bad environment for human beings, EUC sends some robots to explore the treasure. 
To make it easy, we use a graph, which is formed by N points (these N points are numbered from 1 to N), to represent the places to be explored. And some points are connected by one-way road, which means that, through the road, a robot can only move from one end to the other end, but cannot move back. For some unknown reasons, there is no circle in this graph. The robots can be sent to any point from Earth by rockets. After landing, the robot can visit some points through the roads, and it can choose some points, which are on its roads, to explore. You should notice that the roads of two different robots may contain some same point. 
For financial reason, EUC wants to use minimal number of robots to explore all the points on Mars. 
As an ICPCer, who has excellent programming skill, can your help EUC?

Input

The input will consist of several test cases. For each test case, two integers N (1 <= N <= 500) and M (0 <= M <= 5000) are given in the first line, indicating the number of points and the number of one-way roads in the graph respectively. Each of the following M lines contains two different integers A and B, indicating there is a one-way from A to B (0 < A, B <= N). The input is terminated by a single line with two zeros.

Output

For each test of the input, print a line containing the least robots needed.

Sample Input

1 0
2 1
1 2
2 0
0 0

Sample Output

1
1
2

题意：

派机器人去火星寻宝，给出一个无环的有向图，机器人可以降落在任何一个点上，再沿着路去其他点探索，我们的任务是计算至少派多少机器人就可以访问到所有的点。有的点可以重复去。

解析：这题粗略一看，有点最小路径覆盖的意思，但是和标准的最小路径覆盖问题不同在于，标准的最小路径覆盖问题是每个点只能走一次，本题的点是可以重复去走了。

但是我们可以转化一下，传递闭包建立新图，转化为标准的路径覆盖。最小路径覆盖 = 图的顶点数 – 最大匹配数。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 550
using namespace std;
int map[maxn][maxn];
int used[maxn];
int link[maxn];
int n, m;

void init(){
    memset(map, 0, sizeof(map));
}

void getmap(){
    while(m--){
        int a, b;
        scanf("%d%d", &a, &b);
        map[a][b] = 1;
    }
}

void floyd(){
    for(int k = 1; k <= n; ++k)
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                map[i][j] = map[i][j] || map[i][k] && map[k][j];
}

bool dfs(int x){
    for(int i = 1; i <= n; ++i){
        if(map[x][i] && !used[i]){
            used[i] = 1;
            if(link[i] == -1 || dfs(link[i])){
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}

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

int main (){
    while(scanf("%d%d", &n, &m), n || m ){
        init();
        getmap();
        floyd();
        int sum = hungary();
        printf("%d\n", n - sum);
    }
    return 0;
}