POJ1651:Multiplication Puzzle(区间DP)

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动态规划（DP）

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本文介绍了一种使用区间动态规划（DP）算法解决乘法谜题的方法，旨在找到从给定数字序列中取出数字并使总得分最小的策略。文章详细解释了算法流程，并提供了AC代码示例。

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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
10150 + 50205 + 10505 = 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
15020 + 1205 + 1015 = 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
3650

题意
给定一组数a[n]，每次从中取一个数a[i]，算出a[i]*a[i-1]*a[i+1]的和的最小值。

思路
套用区间DP求解。

AC代码

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
int a[110],b[110];
int dp[110][110];
int main()
{
    int n,j;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
        memset(dp,0,sizeof(dp));
    for(int len=2;len<=n-1;len++)
    {
        for(int i=2;i+len<=n+1;i++)
        {
            j=i+len-1;
            dp[i][j]=99999999;
            for(int k=i;k<j;k++)
            dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+a[i-1]*a[k]*a[j]);
        }
    }
    cout<<dp[2][n]<<endl;
    return 0;
}