POJ 3660 floyd 算法

         是一道floyd变形的题目。题目让确定有几个人的位置是确定的，如果一个点有x个点能到达此点，从该点出发能到达y个点，若x+y=n-1，则该点的位置是确定的。用floyd算发出每两个点之间的距离，最后统计时，若dis[a][b]之间无路且dis[b][a]之间无路，则该点位置不能确定。最后用点个数减去不能确定点的个数即可。题目：

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4813 Accepted: 2567

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 54 34 23 21 22 5

Sample Output

2

ac代码：

#include <iostream>#include <string.h>#include <cstdio>using namespace std;#define MAX 0x7fffffffconst int N=110;int dis[N][N];int n,m;void floyd(){for(int k=1;k<=n;++k){for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){if(dis[i][k]!=MAX&&dis[k][j]!=MAX&&dis[i][j]>dis[i][k]+dis[k][j]){  dis[i][j]=dis[i][k]+dis[k][j];}}}}}int main(){//freopen("1.txt","r",stdin);while(~scanf("%d%d",&n,&m)){for(int i=1;i<=n;++i){  for(int j=1;j<=n;++j)  dis[i][j]=MAX;}  int x,y;  while(m--){   scanf("%d%d",&x,&y);   dis[x][y]=1;  }  floyd();  int ans=0;  for(int i=1;i<=n;++i){  for(int j=1;j<=n;++j){if(j==i)continue;    if(dis[i][j]==MAX&&dis[j][i]==MAX){ans++;break;}  }  }  printf("%d\n",n-ans);}  return 0;}