Cow Contest DescriptionN (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.…
题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13085   Accepted: 7289 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kn…
题目链接:https://cn.vjudge.net/problem/POJ-3660 题意 有n头牛,每头牛都有一定的能力值,能力值高的牛一定可以打败能力值低的牛 现给出几头牛的能力值相对高低 问在一场一对一的比赛中,那些牛的排名可以确定下来 思路 一开始还以为是topo排序,每次去掉没有入度或出度的节点 若有两个及以上的节点可以去掉,则排序结束 然后写出来WA两发... 正确思路: 若满足x头牛可以打败牛a,牛a可以打败y头牛,且n==x+y-1时牛a排名唯一确定 那么可以利用Floyd传递…
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: 16941   Accepted: 9447 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 k…
  Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17797   Accepted: 9893 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 oth…
链接 Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu 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. Eac…
Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5989   Accepted: 3234 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…
解题思路:给出n头牛,和这n头牛之间的m场比赛结果,问最后能知道多少头牛的排名. 首先考虑排名怎么想,如果知道一头牛打败了a头牛,以及b头牛打赢了这头牛,那么当且仅当a+b+1=n时可以知道排名,即为此时该牛排第b+1名. 即推出当一个点的出度和入度的和等于n-1的时候,该点的排名是可以确定的, 即用传递闭包来求两点的连通性,如果d[i][j]==1,那么表示i,j两点相连通,度数都分别加1 Cow Contest Time Limit: 1000MS   Memory Limit: 65536…
/* floyd 传递闭包 开始Floyd 之后统计每个点能到的或能到这个点的 也就是他能和几个人确定胜负关系 第一批要有n-1个 然后每次减掉上一批的人数 麻烦的很 复杂度上天了.... 正难则反 我们考虑一定不能确定排名的 */ #include<iostream> #include<cstdio> #include<cstring> #define maxn 110 using namespace std; int n,m,f[maxn][maxn],ans; i…