Treasure Exploration （最小相交覆盖路径二分匹配）-优快云博客

一家名为EUC的公司在火星上寻找宝藏，利用机器人探索无环图中的所有点。为降低成本，需要最少数量的机器人完成任务。输入包含点数和边数，输出最少的机器人数量。解题策略涉及最小相交覆盖路径和二分匹配算法。

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Treasure Exploration

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

Sample Output

1
1
2

这道题跟之前做的最小路径覆盖基本一样，只不过，之前的是最小不相交覆盖路径，即每条路径不会有相同的顶点。所以在求最小相交覆盖路径的时候，要加一个传递闭包，得到任意两个点之间的关系，如果两点之间有路径，那就直接这两点连在一起。接下来就是正常的二分匹配了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define mmm(a,b) memset(a,b,sizeof(a))
const int N=550;
int book[N];
int e[N][N];
int match[N];
int t,n,m;
int dfs(int u)
{
    for(int i=1; i<=n; i++)
    {
        if(e[u][i]&&!book[i])
        {
            book[i]=1;
            if(!match[i]||dfs(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int a,b;
    while(~scanf("%d%d",&n,&m)&&(m+n))
    {
        mmm(e,0);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            e[a][b]=1;
        }
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                for(int k=1; k<=n; ++k)
                    if(e[i][k] && e[k][j])//传递可达性
                        e[i][j] = 1;
        mmm(match,0);
        int ans=0;
        for(int i=1; i<=n; i++)
        {
            mmm(book,0);
            if(dfs(i)) ans++;
        }
        printf("%d\n",n-ans);
    }
}