区间dp模板 石子合并

【2】python算法练习--动态规划专题(1) 区间dp模板题:合并石子 拿到这道题,根据题目所说,需要我们寻找最小合并的数值,很容易就联想到dp,去寻找子问题,化整为零。因为如果直接暴力的话,就是N的阶乘的时间复杂度,肯定没有办法在规定时间内解决问题。 那我们就想:怎么样才会是最小值呢,在分析最小子问题之前,我们要明白一点就是:最后一次合并一定是左边连续的一部分和右边连续的一部分进行合并。 有了这个思路,就很好的可以分解成一个一个小的子问题了,也就是从最小的两个石子开始,相互比较寻找最小值到最后连续的两部分就可以进行最后的运算。 n. 阅读详情

石子合并

设有N堆石子排成一排,其编号为1,2,3,…,N。

每堆石子有一定的质量,可以用一个整数来描述,现在要将这N堆石子合并成为一堆。

每次只能合并相邻的两堆,合并的代价为这两堆石子的质量之和,合并后与这两堆石子相邻的石子将和新堆相邻,合并时由于选择的顺序不同,合并的总代价也不相同。

例如有4堆石子分别为 1 3 5 2, 我们可以先合并1、2堆,代价为4,得到4 5 2, 又合并 1,2堆,代价为9,得到9 2 ,再合并得到11,总代价为4+9+11=24;

如果第二步是先合并2,3堆,则代价为7,得到4 7,最后一次合并代价为11,总代价为4+7+11=22。

问题是:找出一种合理的方法,使总的代价最小,输出最小代价。

输入格式

第一行一个数N表示石子的堆数N。

第二行N个数,表示每堆石子的质量(均不超过1000)。

输出格式

输出一个整数,表示最小代价。

数据范围

1≤N≤300

输入样例:

4
1 3 5 2

输出样例:

22

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN 
洛谷P1880 [NOI1995]石子合并(区间dp模板) 题目描述 在一个圆形操场的四周摆放N堆石子,现要将石子有次序地合并成一堆.规定每次只能选相邻的2堆合并成新的一堆,并将新的一堆的石子数,记为该次合并的得分。 试设计出1个算法,计算出将N堆石子合并成1堆的最小得分和最大得分. 输入格式 数据的第1行试正整数N,1≤N≤100,表示有N堆石子.第2行有N个数,分别表示每堆石子的个数. 输出格式 输出共2行,第1行为最小得分,第2行为最大得... 阅读详情

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