poj 3660 Cow Contest
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8987 | Accepted: 5046 |
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
可以说得上是最短路吧，反正我会是用弗洛伊德暴力水过的，代码很简单，思路清晰就好。
题意：牛之间有绝对的强弱，给出一些胜负关系，问有多少头牛可以确定其绝对排名。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int main() 6 { 7 int n,m,i,j,k; 8 while(~scanf("%d%d",&n,&m)) 9 { 10 int a,b,map[105][105],t; 11 memset(map,0,sizeof(map)); 12 for(i=1; i<=m; i++) 13 { 14 scanf("%d%d",&a,&b); 15 map[a][b]=1; 16 } 17 for(i=1; i<=n; i++) 18 for(j=1; j<=n; j++) 19 for(k=1; k<=n; k++) 20 if(map[j][i]&&map[i][k]) //暴力弗洛伊德，所有的关系连成图 21 map[j][k]=1; 22 int ans=0; 23 for(i=1; i<=n; i++) 24 { 25 int t=0; 26 for(j=1; j<=n; j++) 27 t+=map[i][j]+map[j][i]; 28 if(t==n-1) 29 ans++; 30 } 31 printf("%d ",ans); 32 } 33 return 0; 34 }