想找原题请点击这里:传送门 原题: 题目背景 [Usaco2008 Jan] 题目描述 N ( ≤ N ≤ ) cows, conveniently numbered ..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…

对于一个能够确定名次的点,可以注意到,对于该点,入度和出度的数量加起来等于N-1(这样还是不够准确的确切的说是,能够到达这个点的数量和这个点能够到达的数量的和 floyd不仅可以求两个点之间的最短路径,还能求两个点彼此是否能够相互到达最后对于一个可以确定名次的点,能够到达的所有的点 加上 能够到达该点的所有点的和必须等于n-1当然,我们可以通过二进制来简化这个过程 最后处理结果时,设一个变量flag, 因为该点能够到达本身,flag初值赋为1对于两个点i, j首先f[i][j] | f[j][i…

题目背景 [Usaco2008 Jan] 题目描述 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 competito…

https://www.luogu.org/problem/show?pid=2419 题目背景 [Usaco2008 Jan] 题目描述 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 ski…

P2419 [USACO08JAN]牛大赛Cow Contest Floyd不仅可以算最短路,还可以处理点之间的关系. 跑一遍Floyd,处理出每个点之间是否有直接或间接的关系. 如果某个点和其他$n-1$个点都有关系,那么它的排名就是可确定的. #include<iostream> #include<cstdio> #include<cstring> #define re register using namespace std; ][],ans; int main(…

P2419 [USACO08JAN]牛大赛Cow Contest 题目背景 [Usaco2008 Jan] 题目描述 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 t…

题目背景 [Usaco2008 Jan] 题目描述 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 competito…

题目描述 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,每一轮是两头指定编号的奶牛的对决.如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ,那么她们的对决中,编号为A的奶牛总是能胜出. FJ想知道奶牛们编程能力的…

本蒟蒻又来发题解了, 一道较水的模拟题. 题意不过多解释, 思路如下: 在最开始的时候求出每头牛在t秒的位置(最终位置 然后,如果后一头牛追上了前一头牛,那就无视它, 把它们看成一个整体. else 就++ ans: 上代码: #include<bits/stdc++.h> using namespace std; //要开long long long long n, t, ans = 1, last[100010]; struct node { long long s, p; }a[1000…

题目链接: https://www.luogu.org/problemnew/show/P2419 分析: "在交际网络中,给定若干元素和若干对二元关系,且关系具有传递性. 通过传递性推导出尽量多元素之间的关系的问题叫做传递丢包" --<算法竞赛进阶指南> 所以这道题就用传递丢包来做,怎么实现呢?用Floyd \(f[x][y]\)表示\(x>y\)的关系 最后判断一下对于一个元素\(x\),是不是其他\(n-1\)个元素都与它有传递关系,如果是的话,那么它的位置自然…