传送门 f[i][S] 表示当前集合为 S,最后一个数为 i 的最优解 f[i][S] += f[j][S - i] (j, i ∈ S && j != i && abs(a[i] - a[j]) > k) ——代码 #include <cstdio> #include <iostream> #define LL long long ]; int n, m, k; LL ans, f[][ << ]; inline int read…

P2915 [USACO08NOV]奶牛混合起来Mixed Up Cows 题意: 给定一个长\(N\)的序列,求满足任意两个相邻元素之间的绝对值之差不超过\(K\)的这个序列的排列有多少个? 范围: \(0<=n<=16,0<=序列元素<=25000,0<=k<=3400\) 统计次数一般是递推干的事情,但是我们发现,这个递推并没有一个很明显的顺序关系,并不可以说前几个转移到下一个之类的. 看看数据这么小,一般都是状压干的事情了. 我们可以按照规模进行递推,即一个大小…

P2915 [USACO08NOV]奶牛混合起来Mixed Up Cows 题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large le…

P2915 [USACO08NOV]奶牛混合起来Mixed Up Cows 题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large le…

题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large letters on a gold plate hung around her…

题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large letters on a gold plate hung around her…

题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large letters on a gold plate hung around her…

题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i <= 25,000). The cows are so proud of it that each one now wears her number in a gangsta manner engraved in large letters on a gold plate hung around her…

题目描述 约翰家有N头奶牛,第i头奶牛的编号是Si,每头奶牛的编号都是唯一的.这些奶牛最近 在闹脾气,为表达不满的情绪,她们在挤奶的时候一定要排成混乱的队伍.在一只混乱的队 伍中,相邻奶牛的编号之差均超过K.比如当K = 1时,1, 3, 5, 2, 6, 4就是一支混乱的队伍, 而1, 3, 6, 5, 2, 4不是,因为6和5只差1.请数一数,有多少种队形是混乱的呢? 状压Dp 见代码: #include<cstdio> #include<cstring> #include&l…

类似于n皇后的思想,只要把dfs表示放置情况的数字压缩成一个整数,就能实现记忆化搜索了. 一些有关集合的操作: {i}在集合S内:S&(1<<i)==1: 将{i}加入集合S:S=S|(1<<i): 集合S内包含了{0,1,2,...,n-2,n-1}:S==(1<<n)-1: #include <cstdio> #include <cstring> #include <cmath> using namespace std;…