poj 3660 Cow Contest(warshall算法)
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poj 3660 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:一群牛比赛,每场两只牛对打,并分出胜负,现在问你能确定几只牛的战斗力排名。
解题思路:warshall算法。
#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;typedef long long ll;const int N = 105;int G[N][N];int n, m, t;void floyd() { for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { G[i][j] = G[i][j] || (G[i][k] & G[k][j]); } } }}int main() { while (scanf("%d %d", &n, &m) != EOF) { memset(G, 0, sizeof(G)); int a, b; for (int i = 0; i < m; i++) { scanf("%d%d", &a, &b); G[a][b] = 1; } floyd(); int ans = 0; for (int i = 1; i <= n; i++) { int cnt = 0; for (int j = 1; j <= n; j++) { if (G[i][j]) cnt++; if (G[i][j]) cnt++; } if (cnt == n - 1) ans++; } printf("%d\n", ans); } return 0;}
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