POJ3660——Cow Contest(Floyd+传递闭包)

Cow Contest

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

题目大意：

　　　　给一些牛的排名关系，问有多少牛的排名确定。

解题思路：

　　　　使用Floyd算法来判断传递闭包。

　　　　首先通过输入信息建立邻接矩阵，再使用Floyd求出最短路径Edge[i][j]。

　　　　这时，相对于牛i若Edge[i][j]存在，则说明i肯定打不过j。若Edge[j][i]存在则说明i肯定打得过牛j。

　　　　若i肯定能打过的牛和肯定打不过的牛的和等于牛的总和N-1，则牛i的位置确定。

　　　　据说这个东西叫做传递闭包--！

Code：

 1 #include<stdio.h>
 2 #include<string>
 3 #include<iostream>
 4 #define MAXN 300
 5 using namespace std;
 6 int edge[MAXN+10][MAXN+10];
 7 int N,M;
 8 void init()
 9 {
10     for (int i=1; i<=N; i++)
11         for (int j=1; j<=N; j++)
12             edge[i][j]=INT_MAX;
13 }
14 void floyd()
15 {
16     int m,i,j;
17     for (m=1; m<=N; m++)
18         for (i=1; i<=N; i++)
19             for (j=1; j<=N; j++)
20             {
21                 if (edge[i][m]!=INT_MAX&&edge[m][j]!=INT_MAX&&edge[i][j]>edge[i][m]+edge[m][j])
22                     edge[i][j]=edge[i][m]+edge[m][j];
23             }
24 }
25 int main()
26 {
27     while (cin>>N>>M)
28     {
29         init();
30         for (int i=1; i<=M; i++)
31         {
32             int x1,x2;
33             scanf("%d %d",&x1,&x2);
34             edge[x1][x2]=1;
35         }
36         floyd();
37         int sum=0;
38         for (int i=1; i<=N; i++)
39         {
40             int cnt=0;
41             for (int j=1; j<=N; j++)
42             {
43                 if (i!=j&&edge[i][j]!=INT_MAX)
44                     cnt++;
45                 if (i!=j&&edge[j][i]!=INT_MAX)
46                     cnt++;
47             }
48             if (cnt==N-1) sum++;
49         }
50         printf("%d
",sum);
51     }
52     return 0;
53 }