Multiplication Puzzle （区间dp）

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

3650

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题意：

给出一堆数字，每次可以移除一个数字（首尾数字不能移除），移除一个数字可以得到的分数为相邻两个数字以及他本身这三个数的乘积，最终会剩下首尾两个数字。问可以得到的最小分数为多少？

分析：

明显的区间dp
定义状态 $dp[i][j]$ ：将第$i$个至第$j$个中间的数字全部取走可以得到的最小分数
枚举在区间$(i,j)$最后一个被取走的数字为第$k$个，则有转移方程

$$dp[i][j] = min(dp[i][j] , dp[i][k]+dp[k][j]+a[i]*a[k]*a[j])$$

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<time.h>
#include<iostream>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>

using namespace std;

#define bll long long
const int maxn = 110;
int n,a[maxn],dp[maxn][maxn];

int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        for (int i=1;i<=n;i++) scanf("%d",&a[i]);
        memset(dp,0x3f,sizeof(dp));
        for (int l=2;l<n;l++)
        {
            for (int i=1;i<=n-l;i++)
            {
                int j = i+l;
                for (int k=i+1;k<=j-1;k++)
                {
                    int tmp = a[i]*a[j]*a[k];
                    if (k > i+1) tmp += dp[i][k];
                    if (j > k+1) tmp += dp[k][j];
                    dp[i][j] = min(dp[i][j],tmp);
                }
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}