POJ 1651 Multiplication Puzzle （区间DP）

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

题解： 设dp[i][[j] 为从i到j取完出了a[i]和a[j]这两个数的最小代价。最终所要求出的是dp[1]
。

假设在i到j区间内最后一个取a[k]，那么子问题便可以看得出来：求出dp[i][k]和dp[k][j]的最小代价，求出这两个子问题的代价再加上最后一个取出a[k]的代价，即为dp[i][j]的代价。

问题转化为求出子问题的代价，直到只剩下三个数为止，三个数只能取中间的数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef long long LL;
const int MAX=0x3f3f3f3f;
const int maxn = 100+10;
int n, dp[maxn][maxn], a[maxn];
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 2; i <= n-1; i++) dp[i-1][i+1] = a[i-1]*a[i]*a[i+1];  //边界，三个数只能取中间的数
for(int len = 4; len <= n; len ++)
for(int i = 1; i <= n-len+1; i++) {
int j = i+len-1;
dp[i][j] = MAX;
for(int k = i+1; k <= j-1; k++)
dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j] + a[j]*a[k]*a[i]);
}
printf("%d\n", dp[1]
);
return 0;
}