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POJ - 3660 Cow Contest (floyd)

2017-09-22 16:04 309 查看
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

#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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