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

Source

USACO 2008 January Silver 思路：這道題用的是拓?fù)浣▓D，用佛洛依德建立每個(gè)點(diǎn)與其他點(diǎn)的關(guān)系。如果能確定這個(gè)點(diǎn)與其他所有點(diǎn)的關(guān)系，那它就是可以明確排名的點(diǎn)，否則就無法建立關(guān)系——無法確定排名（自己畫圖）。 #include <cstdio> #include <iostream> #include <string> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std; #define ll long longconst int inf = 0x3f3f3f3f; int n, m, dis[100+8][100+8];int main() {int a, b;memset(dis, 0, sizeof(dis));scanf("%d%d", &n, &m);for(int i = 0; i<m; i++){scanf("%d%d", &a, &b);dis[a][b] = 1;dis[b][a] = -1;}for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)for(int k = 1; k <= n; k++)if(dis[j][i] == dis[i][k] && dis[j][i]){dis[j][k] = dis[i][k];}int ans = 0, sum;for(int i = 1; i <= n; i++){sum = 0;for(int j = 1; j <= n; j++)if(dis[i][j] != 0)sum++;if(sum == n-1)ans++;}printf("%d\n", ans);return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/RootVount/p/11209617.html

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