poj-3660 Cow Contest（传递闭包floyed算法）

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

Sample Output

说明

Cow 2 loses to cows 1, 3, and 4. Thus, cow 2 is no better than any of the cows 1, 3, and 4. Cow 5 loses to cow 2, so cow 2 is better than cow 5. Thus, cow 2 must be fourth, and cow 5 must be fifth. The ranks of the other cows cannot be determined.

题意：有N头牛，评以N个等级，各不相同，先给出部分牛的等级的高低关系，问最多能确定多少头牛的等级

解题思路：

一头牛的等级，当且仅当它与其它N-1头牛的关系确定时确定。

于是我们可以将牛的等级关系看做一张图，然后进行适当的松弛操作（使用floyed求一下传递闭包），得到任意两点的关系，再对每一头牛进行检查即可。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <queue>
 8 const int INF=0x3f3f3f3f;
 9 using namespace std;
10 int D[101][101];
11 
12 int main()
13 {
14     int n,m;
15     scanf("%d %d",&n,&m);
16     //输入信息 
17     for(int i=1;i<=m;i++)
18     {
19         int a,b;
20         scanf("%d %d",&a,&b);
21         D[a][b]=1;//a等级大于b 
22         D[b][a]=-1;//b等级小于a 
23     }
24     
25     //进行适当的松弛操作（使用floyed求一下传递闭包）
26     for(int k=1;k<=n;k++)
27     {
28         for(int i=1;i<=n;i++)
29         {
30             for(int g=1;g<=n;g++)
31             {
32                 if(D[i][k]==1&&D[k][g]==1)
33                     D[i][g]=1;
34                 else if(D[i][k]==-1&&D[k][g]==-1)
35                     D[i][g]=-1;
36                     
37             }
38         }
39     }
40     
41     //统计答案 
42     int ans=0;
43     for(int i=1;i<=n;i++)
44     {
45         int g;
46         for(g=1;g<=n;g++)
47         {
48             if(i!=g&&!D[i][g])//除i=g时，D[i][g]为0说明i对g无确定的关系 
49                 break;
50         }
51         if(g==n+1)//说明该牛对其余的牛都有确定的关系 
52             ans++;
53     }
54     printf("%d\n",ans);
55     return 0;
56 }