poj 1651 Multiplication Puzzle -- (动态规划，区间dp)

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8291		Accepted: 5155

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

今天在算法课上，学到了矩阵连乘的问题，求最少的数乘次数

题意是：一组数字，头和尾都不可以动，每次取出一个数字，然后让这个数和它相邻的数字乘积为其价值，然后相加起来，求最小值。

思路：区间dp，

   动态转移方程 

    k为矩阵链断开的位置

    d数组存取矩阵链计算的最优值，d[i][j]是以第i个矩阵为首，第j个矩阵为尾的     矩阵链的最优值；

     m数组内存放矩阵链的行列信息，m[i-1]和m[i]分别为第i个矩阵的行和列

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int n,p[105];
int dp[105][105];
int main(){
	int i,j,k;
	while(~scanf("%d",&n)){
		for(i=0;i<n;i++)
			scanf("%d",&p[i]);
		for(i=0;i<n;i++)
			dp[i][i]=0;
		int l; //区间长度
		for(l=1;l<n;l++){
			for(i=1,j=l+i;j<n;i++,j++){
				dp[i][j]=inf;
				for(k=i;k<j;k++){
					dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+p[i-1]*p[k]*p[j]);
				}
			}

		}
		printf("%d\n",dp[1][n-1]);
	}
	return 0;
}