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 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 Source |
本题思路：将题目给出的已知边都存入图中，利用传递性求出可能存在的每条边，对于一个学生，用i -> j表示 i 比 j 强，那么对于所有学生，他的排名被确定的条件就是确定他与其它所有同学的排名情况，即Bigger[ i ] + Smaller[ i ] == n - 1。
参考代码：(不建议看，按照上面的思路实现一波就行了)
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 using namespace std;
5
6 const int maxn = 100 + 2, INF = 0x3f3f3f3f;
7 int n, m, a, b;
8 bool G[maxn][maxn];
9
10 int Floyd_Warshall() {
11 int num = 0;
12 for(int k = 1; k <= n; k ++) {
13 for(int i = 1; i <= n; i ++) {
14 for(int j = 1; j <= n; j ++) {
15 G[i][j] = G[i][j] || (G[i][k] && G[k][j]);
16 }
17 }
18 }
19 for(int i = 1; i <= n; i ++) {
20 int temp = 0;
21 for(int j = 1; j <= n; j ++)
22 if(G[i][j] || G[j][i]) temp ++;
23 if(temp == n - 1) num ++;
24 }
25 return num;
26 }
27
28 int main () {
29 scanf("%d %d", &n, &m);
30 int x, y;
31 for(int i = 0; i < m; i ++) {
32 scanf("%d %d", &x, &y);
33 G[x][y] = true;
34 }
35 int ans = Floyd_Warshall();
36 printf("%d\n", ans);
37 return 0;
38 }
原文地址：https://www.cnblogs.com/bianjunting/p/10705863.html