POJ-1651（区间dp）

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本文探讨了一种矩阵连乘的变体问题，即在给定的一串数字中，如何选择取出数字的顺序以使所有取出数字与其两边数字的乘积之和最小。通过动态规划方法，详细解释了算法实现过程。

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题目：http://poj.org/problem?id=1651

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

题意：给你一串数字，从中任意拿出数字（不能拿第一个和最后一个数字），最后只剩下两个数字，求怎样使得所有拿出的数字和其两边的数字的乘机和最小，输出最小值。

分析：矩阵连乘变异。

/*
dp[i][j] 表示i-> j的乘积和的最小值。
dp[i][k] 表示i -> k 的最小值
dp[k][j] 表示k -> j的最小值
无边界
*/

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

const int maxn = 105;
const int INF = 0x3f3f3f3f;
int dp[maxn][maxn],card[maxn];


int main()
{
    int n;
    while(cin >> n)
    {
        for(int i = 0;i < n;i ++)
            cin >> card[i];
        for(int p = 2;p < n;p ++)
            for(int i = 0;i < n - p;i ++)
        {
            int j = i + p;
            dp[i][j] = INF;
            for(int k = i + 1;k < j;k ++)
            {
                dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j] + card[i] * card[k] * card[j]);
            }
        }
        cout << dp[0][n - 1] << endl;
    }
    return 0;
}