當(dāng)前位置：首頁(yè) > 编程资源 > 编程问答 >内容正文

编程问答

nyoj 211 （Floyd算法求传递闭包）

發(fā)布時(shí)間：2025/3/16 编程问答 26 豆豆

生活随笔收集整理的這篇文章主要介紹了 nyoj 211 （Floyd算法求传递闭包）小編覺(jué)得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.

Cow Contest

時(shí)間限制：1000?ms ?|? 內(nèi)存限制：65535?KB 難度：4 描述

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.

輸入

* 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

There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.

輸出

For every case:
* Line 1: A single integer representing the number of cows whose ranks can be determined

樣例輸入

5 5 4 3 4 2 3 2 1 2 2 5 0 0

樣例輸出

解題思路：這里用到了Floyd算法，找到了每個(gè)點(diǎn)與其他點(diǎn)之間的關(guān)系，就能夠確定該點(diǎn)與其他點(diǎn)關(guān)系的個(gè)數(shù)，如果等于n-1就說(shuō)明該點(diǎn)的順序是確定的。。。

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 110; int n,m; bool map[maxn][maxn];int main() { while(scanf("%d%d",&n,&m)!=EOF && m + n){memset(map,false,sizeof(map));int a,b;for(int i = 1; i <= m; i++){scanf("%d%d",&a,&b);map[a][b] = true;}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] = true;int ans = 0, cnt = 0;for(int k = 1; k <= n; k++){ cnt = 0;for(int i = 1; i <= n; i++){if(i == k) continue;if(map[i][k]) cnt++;if(map[k][i]) cnt++;}if(cnt == n-1) ans++;}printf("%d\n",ans);}return 0; }

總結(jié)

以上是生活随笔為你收集整理的nyoj 211 （Floyd算法求传递闭包）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。