poj 1651 Multiplication Puzzle

Multiplication Puzzle

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

题意：给出一列数，需要把中间数子逐个提取出来，取的时候消耗能量是该数字和两边数字的积，例如样例取出1，消耗是10*1*50，最后求提取除头尾的中间所有数字消耗能量之和最小是多少。

用到了区间dp，来分区间进行动态规划，三层循环，最外层是划定区间范围大小，第二层确定边界，最里面是在边界中遍历，k为最后取的数子。

/**区间DP*/
#include 
#include 
#include 
using namespace std;
#define INF 0xfffffff
#define MAXN 110
int dp[MAXN][MAXN], a[MAXN];
int main ()
{
	int n;
	while(~scanf ("%d", &n))
	{
		for (int i = 0; i < n; i++)
			scanf ("%d", &a[i]);
		memset(dp,0,sizeof(dp));
		for (int len = 2; len < n; len++)//len是区间包含的长度减一
		{
			for(int i =0; i