D - Multiplication Puzzle

D - Multiplication Puzzle

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

610 1 50 50 20 5

Sample Output

3650

很多人都说这道题是矩阵连乘，但是我不是很清楚矩阵连乘是怎样的方法，所以就还是当做区间dp把。题意是要把除了两端的剩下的牌一张一张的抽出，每抽出一张就要加上当前牌和他的左右两张牌共3张牌的乘积，要求最后得到的总数最小是多少。

</pre><pre name="code" class="cpp">#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define inf 1e8int main(){    int dp[105][105],a[105],n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        for(int l=3;l<=n;l++)      //需要去掉左右端点，最短长度就是3        {            for(int i=0;i+l-1<n;i++)            {                int j=i+l-1;                dp[i][j]=inf;                for(int k=i+1;k<j;k++)      //枚举中间的位置k                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);            }        }        printf("%d\n",dp[0][n-1]);    }    return 0;}