POJ3660 Cow Contest【最短路-floyd】】的更多相关文章

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…
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.…
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…
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…
  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…
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 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…
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…
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…
poj3660 题意: 有n头牛, 给你m对关系(a, b)表示牛a能打败牛b, 求在给出的这些关系下, 能确定多少牛的排名. 分析: 在这呢先说一下关系闭包: 关系闭包有三种: 自反闭包(r), 对称闭包(s), 传递闭包(t). 先画出 R 的关系图,再画出 r(R), s(R), t(R) 的关系图.                        我们今天用的是传递闭包.   仅作为个人理解 传递闭包: 关系之间具有传递性(例如a> b, b> c, 那么a> c), 在那些已给出…