POJ-1651 Multiplication Puzzle （区间DP）

Multiplication Puzzle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10907 Accepted: 6827

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

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int dp[102][102], a[102];int main(){int n;scanf("%d", &n);for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}memset(dp, 0, sizeof(dp));for(int d = 0; d <= n; ++d){for(int i = 2; i + d < n; ++i){dp[i][i + d] = 1e9;for(int j = i; j <= i + d; ++j){dp[i][i + d] = min(dp[i][i + d], dp[i][j - 1] + dp[j + 1][i + d] + a[i - 1] * a[j] * a[i + d + 1]);}}}printf("%d\n", dp[2][n - 1]);}/*题意：200个数字数列，按一定顺序从中拿走数字，两端的不可以拿，每拿走一个数字的代价是其和相邻两个数字的乘积，现要拿完所有除两端的数字，问最少代价是多少。思路：区间dp，dp[i][j]表示[i,j]内的数字拿完需要多少代价，枚举[i.j]内所有数字作为最后一个拿的数字，转移一下即可。*/