Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9957		Accepted: 6152

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.

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 must contain a single integer - the minimal score.

6
10 1 50 50 20 5

3650
题目的意思是给定一组数据 不能取最左边和最右边的数 然后每次取一个数的代价是取出的数和它周围两边数的乘积 求只剩最左边和最右边两个数的时候 最小代价是多少
首先分解问题了 一般我们定义的时候 dp[i][j]表示i~j 这个区间的子问题 描述的太模糊了
我们定义 dp[i][j]为i~j区间只剩下i,j两个元素需要的代价 这么定义之后  就可以用到区间dp分割区间的思想了
我们架设存在中间值k 是我们一个区间中最后取走的元素 那么就有dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]); 其中k属于i~j
其实就是题意的一个逆向思考 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int inf=10000009;
int main()
{
    int t;
    int dp[101][101],a[101];
    while(cin>>t)
    {
        for(int i=1;i<=t;i++) cin>>a[i];
        //init();
        memset(dp,0,sizeof(dp));
        for(int l=3;l<=t;l++)// 满足题意的最小区间长度为3
        {
            for(int i=1;i+l-1<=t;i++)
            {
                int j=i+l-1;
                dp[i][j]=inf;// 这个初始化比较重要
                for(int k=i+1;k<j;k++)
                {
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
                }
            }
        }
        cout<<dp[1][t]<<endl;
    }
    return 0;
}