POJ 3660 Cow Contest
题目链接:http://poj.org/problem?id=3660
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10066 | Accepted: 5682 |
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:有n头牛,然后有m个条件 每个条件有两个整数a,b表示a能打败b 然后问你有多少头牛能够确定名次。
解题思路:考虑到能够确定名次的牛需要满足一个条件(打败他的牛的个数加上他打败的牛的个数为n-1)
AC代码:
1 #include <stdio.h> 2 #include <string.h> 3 int p[110][110]; 4 int n,m; 5 void floyd() 6 { 7 int i,j,k; 8 for (k = 1; k <= n; k ++) 9 { 10 for (i = 1; i <= n; i ++) 11 { 12 for (j = 1; j <= n; j ++) 13 { 14 if (p[i][k] && p[k][j]) //间接相连也表示能够打败 15 p[i][j] = 1; 16 } 17 } 18 } 19 } 20 int main () 21 { 22 int i,j,a,b; 23 while (~scanf("%d%d",&n,&m)) 24 { 25 memset(p,0,sizeof(p)); 26 27 for (i = 0; i < m; i ++) 28 { 29 scanf("%d%d",&a,&b); 30 p[a][b] = 1; 31 } 32 floyd(); 33 int ans,sum = 0; 34 for (i = 1; i <= n; i ++) 35 { 36 ans = 0; 37 for (j = 1; j <= n; j ++) 38 { 39 ans += p[i][j]; //他打败的牛的个数 40 ans += p[j][i]; //打败他的牛的个数 41 } 42 if (ans == n-1) 43 sum ++; 44 } 45 printf("%d\n",sum); 46 } 47 return 0; 48 }