POJ 1651 Multiplication Puzzle（类似矩阵连乘 区间dp）

传送门：http://poj.org/problem?id=1651

 

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13109		Accepted: 8034

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

Source

Northeastern Europe 2001, Far-Eastern Subregion

 

题目意思：

给你一串数字，头尾不能动，每次取出一个数字，这个数字贡献=该数字与左右相邻数字的乘积，求一个最小值。

分析：

是一个区间dp问题，类似矩阵连乘的做法，也是需要在一个区间中选择一个k值从而来达到你的某种要求，这里是要使得消去的值最小

概况一下题意就是：

初使ans=0,每次消去一个值,位置在pos（pos!=1 && pos !=n）
同时ans+=a[pos-1]*a[pos]*a[pos+1]，一直消元素直到最后剩余2个，求方案最小的ans是多少？

#include <iostream>
#include<algorithm>
#include <cstdio>
#include<cstring>
using namespace std;
#define max_v 105
#define INF 9999999
int a[max_v];
int dp[max_v][max_v];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        for(int r=2;r<n;r++)
        {
            for(int i=2;i<=n-r+1;i++)
            {
                int j=i+r-1;
                int t=INF;
                for(int k=i;k<=j-1;k++)
                {
                    t=min(t,dp[i][k]+dp[k+1][j]+a[i-1]*a[k]*a[j]);
                }
                 dp[i][j]=t;
            }
        }
        printf("%d\n",dp[2][n]);
    }
    return 0;
}