图论---POJ 3660 floyd 算法(模板题)

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: 65536K
Total 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 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

ac代码：

[cpp] view
plaincopy

1. #include <iostream>
2. #include <string.h>
3. #include <cstdio>
4. using namespace std;
5. #define MAX 0x7fffffff
6. const int N=110;
7. int dis[N][N];
8. int n,m;
9. void floyd(){
10. for(int k=1;k<=n;++k){
11. for(int i=1;i<=n;++i){
12. for(int j=1;j<=n;++j){
13. if(dis[i][k]!=MAX&&dis[k][j]!=MAX&&dis[i][j]>dis[i][k]+dis[k][j]){
14. dis[i][j]=dis[i][k]+dis[k][j];
15. }
16. }
17. }
18. }
19. }
20. int main(){
21. //freopen("1.txt","r",stdin);
22. while(~scanf("%d%d",&n,&m)){
23. for(int i=1;i<=n;++i){
24. for(int j=1;j<=n;++j)
25. dis[i][j]=MAX;
26. }
27. int x,y;
28. while(m--){
29. scanf("%d%d",&x,&y);
30. dis[x][y]=1;
31. }
32. floyd();
33. int ans=0;
34. for(int i=1;i<=n;++i){
35. for(int j=1;j<=n;++j){
36. if(j==i)continue;
37. if(dis[i][j]==MAX&&dis[j][i]==MAX)
38. {ans++;break;}
39. }
40. }
41. printf("%d\n",n-ans);
42. }
43. return 0;
44. }

版权声明：本文为博主原创文章，未经博主允许不得转载。