这个状压dp其实很明显,n < 18写在前面了当然是状压.状态其实也很好想,但是有点问题,就是如何判断空间是否够大. 再单开一个g数组,存剩余空间就行了. 题干: 题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after…

状压DP: #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)…

洛谷题目链接:[USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper 题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top…

P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper 题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top…

题目描述 给出n个物品,体积为w[i],现把其分成若干组,要求每组总体积<=W,问最小分组.(n<=18) 输入格式: Line 1: N and W separated by a space. Lines 2..1+N: Line i+1 contains the integer C_i, giving the weight of one of the cows. 输出格式: Line 1: A single integer, R, indicating the minimum number…

题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top of their favorite skyscraper, they had a proble…

题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top of their favorite skyscraper, they had a proble…

题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top of their favorite skyscraper, they had a proble…

题目链接 状压\(dp\) 根据套路,先设\(f[sta]\)为状态为\(sta\)时所用的最小分组数. 可以发现,这个状态不好转移,无法判断是否可以装下新的一个物品.于是再设一个状态\(g[sta]\)表示状态为\(sta\)时 每组剩下的体积的最大值 的最大值,当枚举状态为\(sta\),枚举到第\(i\)个时,可以得到\(g\)的转移: \[ g[v]=max(g[v],g[sta]-w[i]) \] 其中,\(v​\)为转移后的状态. 然后每次就可以根据\(g\)来转移\(f\)了. #…

传送门 输出被阉割了. 只输出最少分的组数即可. f 数组为结构体 f[S].cnt 表示集合 S 最少的分组数 f[S].v 表示集合 S 最少分组数下当前组所用的最少容量 f[S] = min(f[S], f[S - i] + a[i]) (i ∈ S) 运算重载一下即可. ——代码 #include <cstdio> #include <iostream> int n, m, w; ]; struct qwq { int cnt, v; qwq(, ) : cnt(cnt),…