题目链接 Permutation 题目大意:给出n,和m个关系,每个关系为ai必须排在bi的前面,求符合要求的n的全排列的个数. 数据规模为n <= 40,m <= 20. 直接状压DP空间肯定是不够的. 考虑到m <= 20,说明每个连通块的大小不超过21. 那么我们分别对每个连通块求方案数,并且把不同的连通块的方案数组合起来即可. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for…

意甲冠军: 序列p1.p2.p3--pn由1.2.3--n这些数字  现在给出一些条件pi<pj  部条件的排列的个数 思路: 非常easy想到用一条有向的线连接全部的pi和pj  那么就构成了有向无环图(题中说有解所以无环) 又由于pi各不同样  那么题目就变成了有向无环图的拓扑排序的种类数 题目中边数较少  所以可能出现不连通情况  我们先讨论一个连通集合内拓扑排序的种类数 题目中m较小  能够利用状压后的记忆化搜索解决 如今考虑假设知道了A和B两个集合各自的种类数  假设把它们合起来 因为…

题意: 一个有n个数的排列,给你一些位置上数字的大小关系.求合法的排列有多少种. 思路: 数字的大小关系可以看做是一条有向边,这样以每个位置当点,就可以把整个排列当做一张有向图.而且题目保证有解,所以只一张有向无环图.这样子,我们就可以把排列计数的问题转化为一个图的拓扑排序计数问题. 拓扑排序的做法可以参见ZJU1346 . 因为题目中点的数量比较多,所以无法直接用状压DP. 但是题目中的边数较少,所以不是联通的,而一个连通块的点不超过21个,而且不同连通块之间可以看做相互独立的.所以我们可以对…

Permutation Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0. We define the expression [condition] is 1 when condition is True,is 0 whe…

Permutation Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0. We define the expression [condition] is 1 when condition is True,is 0 whe…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3811 Permutation Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 问题描述 In combinatorics a permutation of a set S with N elements is a listing of the elements of S in some…

2019 杭电多校 5 1005 题目链接:HDU 6628 比赛链接:2019 Multi-University Training Contest 5 Problem Description A sequence of length \(n\) is called a permutation if and only if it's composed of the first \(n\) positive integers and each number appears exactly once…

permutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 141    Accepted Submission(s): 81 Problem Description Permutation plays a very important role in Combinatorics. For example ,1 2 3 4…

Permutation Problem Description There is an arrangement of N numbers and a permutation relation that alter one arrangement into another.For example, when N equals to 6 the arrangement is 123456 at first. The replacement relation is 312546 (indicate 1…

这里是一个比较简单的问题:考虑每个数对和的贡献.先考虑数列两端的值,两端的摆放的值总计有2种,比如左端:0,大,小:0,小,大:有1/2的贡献度.右端同理. 中间的书总计有6种可能.小,中,大.其中有两种对答案有贡献,即1/3的贡献度.加和计算可得到答案. Permutation Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s):…