POJ-2594 Treasure Exploration floyd传递闭包+最小路径覆盖，nice！

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8130		Accepted: 3325

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：寻宝，有N个点，m条路，接下来m行每行两个数a,b表示a到b之间连有一条单向路，所以此题图为有向图。每个机器人可以从任意点作为起点然后到达终点，但无法返回，所以求最少需要多少机器人走完这n个点。

思路：很明显最小路径覆盖问题，最小路径覆盖=顶点数-最大点覆盖；但这题的路有点特殊，为了节约成本，当然是只要有路能走下去就一直走下去，题目表明可以重复经过某个点。所以非直接相连的点我们用floyd将其连接，在求连接后的图的最小路径覆盖。这便是此题的思路；

const int N=500+10;
int n,m,w[N][N],linked[N],v[N];
void floyd()//传递闭包
{
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                if(!w[i][j]&&w[i][k]&&w[k][j])
                    w[i][j]=1;
//   for(int k=1; k<=n; k++)
//        for(int i=1; i<=n; i++)
//            printf("%d %d %d\n",k,i,w[k][i]);
}
int dfs(int u)
{
   for(int i=1;i<=n;i++)
       if(w[u][i]&&!v[i])
       {
           v[i]=1;
           if(linked[i]==-1||dfs(linked[i]))
           {
               linked[i]=u;
               return 1;
           }
       }
   return 0;
}
void hungary()
{
    int ans=0;
   memset(linked,-1,sizeof(linked));
   for(int i=1;i<=n;i++)
   {
       memset(v,0,sizeof(v));
       if(dfs(i)) ans++;
   }
   printf("%d\n",n-ans);//最小路径覆盖；
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0) break;
        int u,v;
        memset(w,0,sizeof(w));
        while(m--)
        {
            scanf("%d%d",&u,&v);
            w[u][v]=1;
        }
        floyd();
        hungary();
    }
    return 0;
}

妙哉！