POJ 3660 Cow Contest 传递闭包+Floyd
原题链接:http://poj.org/problem?id=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 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 |
题意
给你若干牛之间的优劣关系,问你有多少头牛能够被确定排名。
题解
如果有x头牛比当前牛弱,有y头牛比当前牛强,并且x+y=n-1,那么这头牛的排名就被唯一确定了。转化为图论问题,我们若牛a比牛b强,则连接a,b(单向)。运用floyd的思想,令dp[i][j]表示从i能够走到j,即牛i和牛j之间存在强弱关系,那么转移就是dp[i][j]=dp[i][j] | (dp[i][k] & dp[k][j]),跑一发floyd,再统计每个点的度即可。
代码
#include<iostream> #include<cstring> #include<vector> #include<queue> #include<algorithm> #define MAX_N 111 using namespace std; bool d[MAX_N][MAX_N]; int n,m; void floyd() { for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) d[i][j] = d[i][j] | (d[i][k] & d[k][j]); } int de[MAX_N]; int main() { cin.sync_with_stdio(false); cin >> n >> m; for (int i = 0; i < m; i++) { int u, v; cin >> u >> v; d[u][v] = 1; } floyd(); int ans = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) de[i] += d[i][j], de[j] += d[i][j]; for (int i = 1; i <= n; i++)if (de[i] == n - 1)ans++; cout << ans << endl; return 0; }