POJ1651:Multiplication Puzzle（简单dp）题解

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探讨了在特定数组条件下，通过策略性选择元素进行乘积操作，以达到最小化总得分的目标。通过动态规划方法解决该问题，实现最优解。

Multiplication Puzzle

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

10 1 50 50 20 5

Sample Output

题意：

给你一串数字，两头的数字不能动，每次划掉一个数字，并将这个数字与其左右两个数字的乘积累加到总价值内，问所有的数字被划掉后（不包括头和尾），总价值的最小值是多少。

思路：不断的计算每个子区间的最小值，最后推广到整个区间。

dp[i][j]表示第i个数到第j个数之间的最小值；

状态转移方程：dp[i][j] =min(dp[i][j], dp[i][k] + dp[k][j] + a[i] * a[j] * a[k] )

#include<cstdio>
#include<cstring> 
int a[105];
int dp[105][105];
int main()  
{  
    int i,j,k,l,t,n;  
    while(scanf("%d",&n)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)  
	    scanf("%d",a+i);  
	    for(l=2;l<n;l++) //l表示子区间的长度 
		{
			for(i=1;i+l<=n;i++)  //i表示起点的那个数 
	        {  
	            j=i+l;  //j表示结束的那个数
	            for(k=i+1;k<j;k++)  //枚举i,j之间所有的数 
	            {  
	                t=dp[i][k]+dp[k][j]+a[i]*a[j]*a[k];  
	                if(!dp[i][j] || dp[i][j]>t)
					{
						dp[i][j]=t;
					}//相当于dp[i][j] =min(dp[i][j], dp[i][k] + dp[k][j] + a[i] * a[j] * a[k]) 
	            }  
	        }  
		}        
	    printf("%d\n",dp[1][n]);  
	} 
    return 0;
}