poj2594（二分图，最小路径覆盖变形）

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7611		Accepted: 3126

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

Sample Output

1
1
2

题意：有很多机器人和1个图，每个机器人从该图的一个点出发，然后现在要求机器人走遍所有的点，求最小的机器人的数量。（2个机器人可以从不同方向走过同一点）。

思路：显然相当于求最路径覆盖嘛，只不过就是最小路径覆盖是每个点不能被走两次的，而现在可以走两次了，所以想一个办法就是把本身不能走的点连起来，比如一个机器人能从a->b->c，可是现在b已经被另一个机器人走过了，最短路径覆盖又不能让一个点被两个人走，可是该机器人必须从a->c才是最优解，所有用floyd将所有点有联系的都连接起来，这样子仍然是原来的最小路径覆盖的问题了。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const int N=555;
bool tu[N][N];
int from[N];///记录右边的点如果配对好了它来自哪里
bool use[N];///记录右边的点是否已经完成了配对
int color[N];
int n,m;
bool dfs(int x)
{for(int i=1; i<=m; i++) ///m是右边，所以这里上界是mif(!use[i]&&tu[x][i]){use[i]=1;if(from[i]==-1||dfs(from[i])){from[i]=x;return 1;}}return 0;
}
int hungary()
{int tot=0;memset(from,-1,sizeof(from));for(int i=1; i<=n; i++) ///n是左边，所以这里上界是n{memset(use,0,sizeof(use));if(dfs(i))tot++;}return tot;
}
int main()
{int k;while(cin>>n>>k&&n+k){m=n;memset(tu,0,sizeof(tu));while(k--){int a,b;cin>>a>>b;tu[a][b]=1;}for(int k=1; k<=n; k++)for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if(tu[i][k]&&tu[k][j])tu[i][j]=1;printf("%d\n",n-hungary());}return 0;
}

转载于:https://www.cnblogs.com/martinue/p/5490459.html

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