POJ 2594 Treasure Exploration（最大路径覆盖）

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 5480		Accepted: 2154

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

Source

POJ Monthly--2005.08.28,Li Haoyuan

 

 

此题跟普通的路径覆盖有点不同。

You should notice that the roads of two different robots may contain some same point.

也就是不同的路径可以有重复点。

先用floyed求闭包，就是把间接相连的点也连起来。

/*

POJ 2594

求最大独立集=顶点数-最大匹配数

*/

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;

/* **************************************************************************

//二分图匹配（匈牙利算法的DFS实现）

//初始化：g[][]两边顶点的划分情况

//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配

//g没有边相连则初始化为0

//uN是匹配左边的顶点数，vN是匹配右边的顶点数

//调用：res=hungary();输出最大匹配数

//优点：适用于稠密图，DFS找增广路，实现简洁易于理解

//时间复杂度:O(VE)

//***************************************************************************/

//顶点编号从0开始的

const int MAXN=510;

int uN,vN;//u,v数目

int g[MAXN][MAXN];

int linker[MAXN];

bool used[MAXN];

bool dfs(int u)//从左边开始找增广路径

{

    int v;

    for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改

      if(g[u][v]&&!used[v])

      {

          used[v]=true;

          if(linker[v]==-1||dfs(linker[v]))

          {//找增广路，反向

              linker[v]=u;

              return true;

          }

      }

    return false;//这个不要忘了，经常忘记这句

}

int hungary()

{

    int res=0;

    int u;

    memset(linker,-1,sizeof(linker));

    for(u=0;u<uN;u++)

    {

        memset(used,0,sizeof(used));

        if(dfs(u)) res++;

    }

    return res;

}

//******************************************************************************/



void floyed(int n)//求传递闭包

{

    for(int i=0;i<n;i++)

      for(int j=0;j<n;j++)

      {

          if(g[i][j]==0)

          {

              for(int k=0;k<n;k++)

              {

                  if(g[i][k]==1&&g[k][j]==1)

                  {

                      g[i][j]=1;

                      break;

                  }

              }

          }

      }

}



int main()

{

    int n,m;

    int u,v;

    while(scanf("%d%d",&n,&m))

    {

        if(n==0&&m==0)break;

        uN=vN=n;

        memset(g,0,sizeof(g));

        while(m--)

        {

            scanf("%d%d",&u,&v);

            u--;v--;

            g[u][v]=1;

        }

        floyed(n);

        printf("%d\n",n-hungary());

    }

    return 0;

}