题目描述 In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For…

题面 传送门:https://www.luogu.org/problemnew/show/P3119 Solution 这题显然要先把缩点做了. 然后我们就可以考虑如何处理走反向边的问题. 像我这样的蒟蒻,当然是使用搜索,带记忆化的那种(滑稽). 考虑设f(i,j)表示到达第i个点,还能走j次反向边,所能到达的最多的点的数量. 转移可以表示为: 如果x能到达1所在的强连通分量或max出来的值不为0,说明当前状态可行,否则不可行. 然后用记忆化搜索表达出来就OK了 Code #include<io…

P3119 [USACO15JAN]草鉴定Grass Cownoisseur 题目描述 约翰有\(n\)块草场,编号1到\(n\),这些草场由若干条单行道相连.奶牛贝西是美味牧草的鉴赏家,她想到达尽可能多的草场去品尝牧草. 贝西总是从1号草场出发,最后回到1号草场.她想经过尽可能多的草场,贝西在通一个草场只吃一次草,所以一个草场可以经过多次.因为草场是单行道连接,这给贝西的品鉴工作带来了很大的不便,贝西想偷偷逆向行走一次,但最多只能有一次逆行.问,贝西最多能吃到多少个草场的牧草. 输入输出格式…

P3119 [USACO15JAN]草鉴定Grass Cownoisseur 题目描述 In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way co…

P3119 [USACO15JAN]草鉴定Grass Cownoisseur 题目描述 In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way co…

题目描述 In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For…

http://www.lydsy.com/JudgeOnline/problem.php?id=3887|| https://www.luogu.org/problem/show?pid=3119 Description In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm con…

https://www.luogu.org/problemnew/show/P3119 题意 有一个有向图,允许最多走一次逆向的路,问从1再走回1,最多能经过几个点. 思路 (一)首先先缩点.自己在缩点再建图中犯了错误,少连接了大点到其他点的边.跑两次最长路,一次以1为起点,一次以1为终点(跑一遍反图)然后枚举边,判断可否形成一个环.(二)分层图的思想以为只有一次逆向的机会,可以建两层图,第一层向第二层连翻转的情况. #include <algorithm> #include <iter…

思路很乱,写个博客理一理. 缩点 + dp. 首先发现把一个环上的边反向是意义不大的,这样子不但不好算,而且相当于浪费了一次反向的机会.反正一个强连通分量里的点绕一遍都可以走到,所以我们缩点之后把一个强连通分量放在一起处理. 设$st$表示缩点之后$1$所在的点,设$f_{x}$表示从$st$走到$x$的最长链,$g_{x}$表示从$x$走到$st$的最长链,因为把一个$DAG$上的边反向一下并不会走重复的点,那么我们最后枚举一下边$(x, y)$,把它反向,这样子$f_{x} + g_{y}…

屠龙宝刀点击就送 Tarjan缩点+拓扑排序 以后缩点后建图看n范围用vector ,或者直接用map+vector 结构体里数据要清空 代码: #include <cstring> #include <cctype> #include <cstdio> #include <vector> #include <queue> #define N 100005 using namespace std; vector<int>G[N]; i…