POJ2594 Treasure Exploration（最小路径覆盖）

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8550		Accepted: 3495

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

————————————————————————————

一个有向图中， 有若干条连接的路线， 问最少放多少个机器人，可以将整个图上的点都走过。 最小路径覆盖问题。

思路：最小路径覆盖问题， 最小路径覆盖 = |V| - 最大匹配数。 当然做这道题还有一个坑！！ 如果有向图的边有相交的情况，那么就不能简单的对原图求二分匹配了，建图时dfs预处理将每个点能到达的点都与他连起来

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;
 
#define LL long long
const int INF = 0x3f3f3f3f;
const int MAXN=1005;
int uN,vN,n;  //u,v数目
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];
int link[MAXN];
int vis[MAXN];
bool dfs(int u)
{
    int v;
    for(v=1; v<=vN; 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 hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1; u<=uN; u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}
 
void dfs2(int pos,int x)
{
    for(int i=0; i<n; i++)
    {
        if(g[pos][i]==1&&!vis[i])
        {
            vis[i]=1;
            g[x][i]=1;
            dfs2(i,x);
        }
    }
}
 
int main()
{
    int m,k,x,y,T;
    while(~scanf("%d%d",&n,&m)&&(m||n))
    {
 
        memset(g,0,sizeof g);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&x,&y);
            g[x][y]=1;
        }
        for(int i=1; i<=n; i++)
        {
            memset(vis,0,sizeof vis);
            vis[i]=1;
            dfs2(i,i);
        }
        uN=vN=n;
        printf("%d\n",n-hungary());
    }
    return 0;
}