During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting. Now, you're given n teams, you…
During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting. Now, you're given n teams, you…
原题 During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting. Now, you're given n teams, y…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 遍历 日期 题目地址:https://leetcode-cn.com/problems/output-contest-matches/ 题目描述 During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, lik…
原题链接在这里:https://leetcode.com/problems/output-contest-matches/description/ 题目: During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good…
During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting. Now, you're given n teams, you…
1612: [Usaco2008 Jan]Cow Contest奶牛的比赛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 645  Solved: 433[Submit][Status] 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. 整个比赛被分成了若干轮,…
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…
对于第 i 头牛 , 假如排名比它高和低的数位 n - 1 , 那么他的 rank 便可以确定 . floyd --------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep…
Cow Contest奶牛的比赛 bzoj-1612 Usaco-2008 Jan 题目大意:题目链接. 注释:略. 想法: 我们对于每个点dfs,看一下比这个点大的点加上比这个点小的点是否是n-1即可. 最后,附上丑陋的代码... ... #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N…