POJ 1651：Multiplication Puzzle 矩阵相乘式DP

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

6
10 1 50 50 20 5

Sample Output

3650

题意就是给了几个数，依次拿走中间的数，每一次拿走第i个数都会有相应的代价value[i-1]*value[i]*value[i+1]，问要求的是给定的序列 拿走所有中间的数 代价最少是多少。

就是矩阵相乘式DP。。。把那个数单提出来想象成要把它第一个拿走就OK了啊，怎么自己的脑筋总是奔到死胡同里面呢，气死我了。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

#define INF 0x3f3f3f3f

int dp[105][105];
int value[105];
int num;

int main()
{
	int i,j,k,len;
	scanf("%d",&num);

	for(i=1;i<=num;i++)
	{
		scanf("%d",&value[i]);
		dp[i][i]=0;
	}

	for(i=1;i<=num;i++)
	{
		dp[i][i+1] = 0;
		dp[i][i+2] = value[i]*value[i+1]*value[i+2];
	}

	for(len=3;len<=num;len++)
	{
		for(i=1,j=len+i;j<=num;i++)
		{
			j=len+i;
			dp[i][j]=INF;
			for(k=i+1;k<j;k++)
			{
				dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+value[k]*value[i]*value[j]);//想象成是第一个拿走的
			}
		}
	}
	cout<<dp[1][num]<<endl;
	return 0;
}

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