When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m dishes. But at that, he doesn't want to order the same dish twice to taste as many dishes as pos…

题意翻译 kefa进入了一家餐厅,这家餐厅中有n个菜(0<n<=18),kefa对第i个菜的满意度为ai(0<=ai<=10^9),并且对于这n个菜有k个规则,如果kefa在吃完第xi个菜之后吃了第yi个菜(保证xi.yi不相等),那么会额外获得ci(0<=ci<=10^9)的满意度.kefa要吃m道任意的菜(0<m<=n),但是他希望自己吃菜的顺序得到的满意度最大,请你帮帮kefa吧! 输入第一行是三个数:n,m,k 第二行是n个数,第i个数表示kefa对…

题目传送门 友情链接:new2zydalao%%%  一篇优秀的状压文章 题目大意:$n$个菜有$k$个规则,如果kefa在吃完第$xi$个菜之后吃了第$yi$个菜(保证$xi$.$yi$不相等), 那么会额外获得$ci$ (0<=$ci$<=$10^9$)(0<=$ci$<=$10^9$)的满意度.(0<$n$<=18).但是他希望自己吃菜的顺序得到的满意度最大, 请你帮帮kefa吧! 数据范围这么小==,应该会用到状压的.但是之前状压都是那些比较明显的在棋盘上的类型…

枚举最后食物 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); a <= (c); ++ a) #define nR(a,b,c) for(register int a = (b); a >= (c); -- a…

题目链接: 题目 D. Kefa and Dishes time limit per test:2 seconds memory limit per test:256 megabytes 问题描述 When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly…

Kefa and Dishes Problem's Link Mean: 菜单上有n道菜,需要点m道.每道菜的美味值为ai. 有k个规则,每个规则:在吃完第xi道菜后接着吃yi可以多获得vi的美味值. 问:最多可以获得多少美味值? (1≤m≤n≤18,0≤k≤n∗(n−1)) analyse: 经典的状压DP. 由于最多18道菜,可用一个数s(s<=2^18)来唯一标识一种状态. 对于一个状态s,枚举两个位置i和j:i从已选择的菜中选定,j从未选择的菜中选定. 下一个状态ss的就是:吃完i后接着…

Description When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m dishes. But at that, he doesn't want to order the same dish twice to taste as many d…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he ne…

题目大概说要吃掉n个食物里m个,吃掉各个食物都会得到一定的满意度,有些食物如果在某些食物之后吃还会增加满意度,问怎么吃满意度最高. dp[S][i]表示已经吃掉的食物集合是S且刚吃的是第i个食物的最大满意度 ..没什么好说的 #include<cstdio> #include<algorithm> using namespace std; ],pairs[][]; <<][]; int getCnt(int s){ ; ; i<; ++i){ ) ++cnt; }…

题意:给定n个菜,每个菜都有一个价值,给定k个规则,每个规则描述吃菜的顺序:i j w,按照先吃i接着吃j,可以多增加w的价值.问如果吃m个菜,最大价值是多大.其中n<=18 思路:一看n这么小,除了暴力之外就得想想状态压缩dp了.因为每种菜正好两种状态(吃过与没吃过),正好可以使用二进制来表示每种状态,那么一共有(1<<n)位种可能,即从状态(000...0)到状态(111...1),所以定义状态dp[s][i],表示状态为s时,并且最后吃第i种菜的时候的最大值.那么转移方程就是 dp…