原题链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8395   Accepted: 4734 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kno…

Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3660 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some co…

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…

Cow Contest POJ - 3660 :http://poj.org/problem?id=3660   参考:https://www.cnblogs.com/kuangbin/p/3140837.html   题意: n头牛,有m对牛进行了比赛,现在告诉你每队牛比赛的结果,A胜B,问有几头牛的排名可以确定. 思路: 题目给出了m对的相对关系,求有多少个排名是确定的. 使用floyed求一下传递闭包.如果这个点和其余的关系都是确定的,那么这个点的排名就是确定的. #include <al…

Cow Contest 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/H 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 certa…

传送门:http://poj.org/problem?id=3660 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, 能确定多少头牛的排名. 传递闭包: 关系之间具有传递性(例如a> b, b> c, 那么a> c), 在那些已给出的关系基础上, 通过传递性, 把所有可能的关系都找出来. 思路:假设一头牛可以被X头牛打败,可以打败Y头牛.如果这头牛的排名可以确定则X+Y=N-1.想到Floyd正好可以求X→Y的关系,只是将Floyd求距离改成了判…

<题目链接> 题目大意: 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. 解题分析: 首先,做这道题要明确,什么叫确定牛的排名.假设该牛被x头牛打败(直接或间接),同时它也有y头手下败将(直接或间接),当x+y=n-1时,即除这头牛本身外,其他所有的牛都为这头牛贡献了出度或者入度.即,当这头牛与其它所有的牛的输赢关系都确定时(直接或间接),这头牛的排名也就可以确定了.而题目只给出了一些牛的直接输赢关系,这时,我们就可以利用Floyed算法…

传送门 解题思路 考试题,想到传递闭包了,写了个O(n^3)的,T了7个点...后来看题解是tm的bitset优化???以前好像没听过诶(我太菜了),其实也不难,时间复杂度O(n^3/32) #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<bitset> using namespace std; ; inline int rd(){ ,f…

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…

题目链接: http://poj.org/problem?id=3660 题目大意: 给出n头牛,m个关系,关系为a的战力比b高.求最后可以确定排名的牛的数量 思路: 1.如果一头牛跟其他所有牛都确定了一个输赢关系,那么该牛的排名就得到了确定,所以用floyd跑一遍传递闭包.然后求得每个点的出度(赢了), 入度(输了).若该点的度之和为 n - 1 ,即确定排名. #include<stdio.h> #include<string.h> #define mem(a, b) mems…