Multiplication Puzzle POJ - 1651（区间dp）_multiplication puzzle poj详解-优快云博客

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/81427611

本文介绍了一道经典的动态规划问题——最小得分乘法谜题。玩家从一组有序整数中移除数字，每次移除的数字得分等于该数字乘以其相邻的两个数字。目标是以特定顺序移除所有数字（除了第一个和最后一个），使得总得分最小。

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Multiplication Puzzle POJ - 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：

就是给出一组排好序列的数，你可以从中去掉数字，每一次去掉的数字就会有一次的得分，得分为去掉的这个数字乘以它前面的那个数字和它后面的那个数字。这样子去数字直到只剩下原数列的第一个数和原数列的最后一个数~求最小得分

分析：

定义 $dp[i][j]$ 表示区间 $[i,j]$ 的最小得分

一开始可以先初始化好长度为三的区间值是多少，可以直接算

然后开始枚举区间长度为4到更长的区间 $[i,j]$

枚举区间内的一个点k作为最后一次取的数（即其他数都已经取走了，只剩 $i,k,j(i < k < j)$ ）

状态转移方程为

$dp[i][j] = dp[i][k] + dp[k][j] + a[i] * a[k] * a[j]$

code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int dp[110][110];
int a[110];
int n;
int main(){
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n; i++){
            dp[i][i+2] = a[i] * a[i+1] * a[i+2];
        }
        for(int len = 4; len <= n; len++){//区间长度
            for(int i = 1; i + len - 1 <= n; i++){//区间左端点
                int j = i + len - 1;
                dp[i][j] = 10000000;
                for(int k = i + 1; k < j; k++){//枚举中间最后一个取的数
                    dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j] + a[i] * a[k] * a[j]);
                }
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}