poj1651 java 动态规划矩阵连乘

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10599		Accepted: 6607

DescriptionThe 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 thecard 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, scoring10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.InputThe 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.OutputOutput must contain a single integer - the minimal score.Sample Input

610 1 50 50 20 5

Sample Output

3650

题意：有n个数，每次取走一个数，并且与左右相乘的和累加（第一个和最后一个数不能取），求最小值。运用矩阵连乘。import java.io.BufferedReader;import java.io.InputStreamReader;import java.util.Scanner;public class Main {@SuppressWarnings("resource")public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (sc.hasNext()) {int n = sc.nextInt();int arr[] = new int[n + 1];for (int i = 1; i < arr.length; i++) {arr[i] = sc.nextInt();}int dp[][] = new int[n + 1][n + 1];for (int len = 2; len < n; len++)for (int l = 1; l + len - 1 <= n; l++) {int r = l + len - 1;dp[l][r] = Integer.MAX_VALUE;for (int k = l; k < r; k++)dp[l][r] = Math.min(dp[l][r], dp[l][k] + dp[k + 1][r] + arr[l - 1] * arr[k] * arr[r]);}System.out.println(dp[2]);}}}