POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest(图论,传递闭包) 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…

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 contes…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16341   Accepted: 9146 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 other…

1612: [Usaco2008 Jan]Cow Contest奶牛的比赛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 891  Solved: 590[Submit][Status][Discuss] Description FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:).在赛场上,奶牛们按1..N依次编号.每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名. 整个比…

题意 : 给出 N 头奶牛在比赛的结果,问你最多的能根据给出结果确定其名次的奶牛头数.结果给出的形式为 A  B 代表在比赛当中 A 战胜了 B 分析 : 对于一头奶牛来说,如果我们能确定其他 N - 1 头奶牛和它的关系,那么它的名次就确定了.将奶牛之间的胜负关系建图,如果给出 A B 那么我们建一条 A->B 的边,代表 A 能战胜 B ,当然也表示了 A 和 B 能确立关系,那么现在有一头奶牛 C 且有 C->A 即其与 A 的关系是确定的,那么 B 和 C 的关系是否能确定呢?毋庸置疑…

题目链接:http://poj.org/problem?id=3660 Description N ( ≤ N ≤ ) cows, conveniently numbered ..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 am…

题目链接:http://poj.org/problem?id=3660 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 uniq…

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 conduct…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…

#include<iostream> #include<cstring> using namespace std; ,INF=0x3f3f3f3f; int f[N][N]; int main() { int n,m; cin>>n>>m; memset(f,0x3f,sizeof f); int x,y; ;i<m;i++) { cin>>x>>y; //x>y f[x][y]=; //x<y f[y][x]=-;…