POJ1651Multiplication Puzzle(区间dp)

最新推荐文章于 2023-12-29 13:09:48 发布

丶先森

最新推荐文章于 2023-12-29 13:09:48 发布

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分类专栏： ACM DP 日常总结

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探讨了一种通过动态规划解决乘法谜题的方法，旨在找出取卡片顺序以使总分最小化的策略。该问题涉及一排含有正整数的卡片，玩家需遵循特定规则移除卡片并计算分数。

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9836		Accepted: 6083

Description

The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.

The goal is to take cards in such order as to minimize the total number of scored points.

For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.

Input

The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.

Output

Output must contain a single integer - the minimal score.

Sample Input

6
10 1 50 50 20 5

Sample Output

Source

Northeastern Europe 2001, Far-Eastern Subregion

题目大意:n个数，然后取n-2次数字，不能取头尾两个数字，然后每取一个数，得分是目前此数的前后两个数的加上这个数的乘积，求出来最小的乘积和是多少

解题思路:这是一道区间DP，主要是dp[j][k]指的是j到k区间的最小乘积是多少，且j和k位置最后抓取

这道题根据我之前写的题来看，发现求最小值的时候一定要给dp区间一个特别大的值，然后在转移方程的时候才能保证min之后不是0

#include<iostream>    
#include<cstdio>  
#include<stdio.h>  
#include<cstring>    
#include<cstdio>    
#include<climits>    
#include<cmath>   
#include<vector>  
#include <bitset>  
#include<algorithm>    
#include <queue>  
#include<map>  
using namespace std;

long long int a[105], dp[105][105], i, n, j, u, k;
int main()
{
	cin >> n;
	for (i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	for (i = 3; i <= n; i++)
	{
		for (j = 1, k = i; k <= n; j++, k++)
		{
			dp[j][k] = 10000000;
			for (u = j+1; u < k; u++)
			{
				dp[j][k] = min(dp[j][k], dp[j][u] + dp[u][k] + a[k] * a[j] * a[u]);
			}
		}
	}
	cout << dp[1][n] << endl;
}