区间dp_3

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 Outpu

3650

每次去牌都只跟左右断电有关，问题的解就是取走最后一张牌的得分+两个子区间上的最小得分。不妨假设当前区间为[i, j],在(i,j)之间枚举最后一张被取走的牌，通过最优子问题求出当前区间的最优解：

dp[i][j] = min{ dp[i][k]+dp[k][j] + a[i]*a[j]*a[k]   (i+1<=k<=j-1) }

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int a[110];int dp[110][110];int main(){    int n;    int len,i,j,k;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%d",&a[i]);    memset(dp,0,sizeof(dp));    for(len=3;len<=n;len++)//len一定要取到等号    {        for(i=0;i<n;i++)        {            j=len+i-1;// 要准确,否则会出错            dp[i][j]=10000000;            for(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]);//关键部分        }    }    printf("%d\n",dp[0][n-1]);    return 0;}