Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 6284		Accepted: 2533

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?

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.

For each test of the input, print a line containing the least robots needed.

1 0
2 1
1 2
2 0
0 0

1
1
2

POJ Monthly--2005.08.28,Li Haoyuan

 1 //1164K    907MS    C++    1019B    2013-11-09 17:45:32
 2 /*
 3
 4    题意：
 5        给你一个图，要求用最少的机器人遍历全部的点
 6
 7    最短路+二分匹配：
 8        这题有点意思，开始WA好几次没想通，就看了下别人的解题报告，果然被坑课..
 9
10        二分图的最小路径覆盖是要求每个点只能通过一次，而此题则是可以通过多次
11        eg.
12            5 5
13            1 3
14            2 3
15            3 4
16            3 5
17        使用二分图的最小路径覆盖得出的答案是：n-最大匹配=3
18                          此题得出的答案应是：n-最大匹配=2
19
20        故在读完图后要再次构图，使用floyd(此题数据较弱)，然后是某些点可以连通
21        在使用求最小路径覆盖即可求出解
22
23 */
24 #include<stdio.h>
25 #include<string.h>
26 int g[505][505];
27 int match[505];
28 int vis[505];
29 int n,m;
30 void floyd()
31 {
32     for(int k=1;k<=n;k++)
33         for(int i=1;i<=n;i++)
34             for(int j=1;j<=n;j++)
35                 if(g[i][k] && g[k][j])
36                     g[i][j]=1;
37 }
38 int dfs(int x)
39 {
40     for(int i=1;i<=n;i++){
41         if(!vis[i] && g[x][i]){
42             vis[i]=1;
43             if(match[i]==-1 || dfs(match[i])){
44                 match[i]=x;
45                 return 1;
46             }
47         }
48     }
49     return 0;
50 }
51 void hungary()
52 {
53     memset(match,-1,sizeof(match));
54     int cnt=0;
55     for(int i=1;i<=n;i++){
56         memset(vis,0,sizeof(vis));
57         if(dfs(i)) cnt++;
58     }
59     printf("%d\n",n-cnt);
60 }
61 int main(void)
62 {
63     int a,b;
64     while(scanf("%d%d",&n,&m),n+m)
65     {
66         memset(g,0,sizeof(g));
67         for(int i=0;i<m;i++){
68             scanf("%d%d",&a,&b);
69             g[a][b]=1;
70         }
71         floyd();
72         hungary();
73     }
74     return 0;
75 }