POJ - 3660 Cow Contest (floyd)
2017-09-22 16:04
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题目大意:给出 n 头牛的强弱关系,问有几头牛能够确定排名
解题思路:为每个关系建立一条边,间接有关的用 floyd 建好,然后统计与剩余 n-1 个点都相连的点的个数
解题思路:为每个关系建立一条边,间接有关的用 floyd 建好,然后统计与剩余 n-1 个点都相连的点的个数
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<algorithm> #include<cmath> #include<string.h> #include<string> #include<queue> #define max(a,b) ((a)>(b)?(a):(b)) #define min(a,b) ((a)<(b)?(a):(b)) const int INF = 0x3f3f3f3f; const int NINF = -INF -1; const int MAXN = 100+5; using namespace std; int n, m; int map[MAXN][MAXN]; int main() { while (scanf("%d%d", &n, &m) != EOF) { memset(map, 0, sizeof(map)); for (int i = 0; i < m; i++) { int a, b; scanf("%d%d", &a, &b); map[a][b] = 1; } for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (map[i][k] && map[k][j]) map[i][j] = 1; int ans = 0; for (int i = 1; i <= n; i++) { int tmp = 0; for (int j = 1; j <= n; j++) if (map[i][j] || map[j][i]) tmp++; if (tmp == n-1) ans++; } printf("%d\n", ans); } return 0; }
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