题目链接:https://www.luogu.org/problemnew/show/P2881 题目链接:https://vjudge.net/problem/POJ-3275 题目大意 给定标号为 1~N 这 N 个数,在给定 M 组大小关系,求还需要知道多少组大小关系才可以给这组数排序? 分析1(Floyd + bitset) 总共需要知道 n * (n - 1) / 2 条边,因此只要求一下现在已经有了多少条边,再减一下即可.由于大小关系有传递性,因此计数之前需要求传递闭包. 直接上 f…

NC25025 [USACO 2007 Nov G]Sunscreen 题目 题目描述 To avoid unsightly burns while tanning, each of the \(C\) (\(1 ≤ C ≤ 2500\)) cows must cover her hide with sunscreen when they're at the beach. Cow \(i\) has a minimum and maximum \(SPF\) rating (\(1 ≤ minS…

1703: [Usaco2007 Mar]Ranking the Cows 奶牛排名 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 323  Solved: 238[Submit][Status][Discuss] Description     农夫约翰有N(1≤N≤1000)头奶牛,每一头奶牛都有一个确定的独一无二的正整数产奶率．约翰想要让这些奶牛按产奶率从高到低排序．    约翰已经比较了M(1≤M≤10000)对奶牛的产奶率,但他发现,他还需…

传送门 题意: 农场主 FJ 有 n 头奶牛,现在给你 m 对关系(x,y)表示奶牛x的产奶速率高于奶牛y: FJ 想按照奶牛的产奶速率由高到低排列这些奶牛,但是这 m 对关系可能不能精确确定这 n 头奶牛的关系: 问最少需要额外增加多少对关系使得可以确定这 n 头奶牛的顺序: 题解: 之所以做这道题,是因为在补CF的题时用到了bitset<>: 搜这个容器的用法是看到了一篇标题为POJ-3275:奶牛排序Ranking the Cows(Floyd.bitset)的文章: 正好拿着道题练练b…

NC24840 [USACO 2009 Mar S]Look Up 题目 题目描述 Farmer John's N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height Hi (1 <= Hi <= 1,000,000). Each cow is looking to her left toward those with hig…

NC25043 [USACO 2007 Jan S]Protecting the Flowers 题目 题目描述 Farmer John went to cut some wood and left \(N (2 ≤ N ≤ 100,000)\) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating hi…

Ranking the cows Description Each of Farmer John's N cows (1 ≤ N ≤ 1,000) produces milk at a different positive rate, and FJ would like to order his cows according to these rates from the fastest milk producer to the slowest. FJ has already compared…

Ranking the Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3301   Accepted: 1511 Description Each of Farmer John's N cows (1 ≤ N ≤ 1,000) produces milk at a different positive rate, and FJ would like to order his cows according to…

2019-08-21便宜的回文(USACO 2007) 内存限制:128 MiB 时间限制:1000 ms 标准输入输出 题目类型:传统 评测方式:文本比较 题目描述 追踪每头奶牛的去向是一件棘手的任务,为此农夫约翰安装了一套自动系统.他在每头牛身上安装了一个电子身份标签,当奶牛通过扫描器的时候,系统可以读取奶牛的身份信息.目前,每个身份都是由一个字符串组成的,长度为M (1≤M≤2000),所有的字符都取自小写的罗马字母. 奶牛们都是顽皮的动物,有时她们会在通过扫描器的时候倒着走,这样一个原来…

POJ3275 Ranking the Cows #include <iostream> #include <cstdio> #include <bitset> using namespace std; ; int n, m; bitset<maxn> maps[maxn]; void floyd() { ; k <= n; k++) { ; i <= n; i++) { if (maps[i][k]) maps[i] |= maps[k]; }…