【算法】Multiplication Puzzle-动态规划

LTC0316

于 2022-11-04 10:28:40 发布

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分类专栏：数据结构与算法文章标签：算法动态规划 c++

本文链接：https://blog.youkuaiyun.com/qq_43062936/article/details/127684065

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本文介绍了一种通过动态规划解决乘法谜题的方法。玩家需要从一排含有正整数的卡片中按最优顺序取出卡片，使得得分最小。文章详细解释了如何使用动态规划求解该问题，并给出了具体的实现代码。

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Multiplication Puzzle

总时间限制: 1000ms 内存限制: 65536kB

描述
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
10150 + 50205 + 10505 = 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
15020 + 1205 + 1015 = 1000+100+50 = 1150.
输入
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.
输出
Output must contain a single integer - the minimal score.
样例输入
6
10 1 50 50 20 5
样例输出
3650

#include <iostream>
#include <climits>
#include <cmath>
using namespace std;

int main() {
    int N;
    cin >> N;
    int array[101];
    int dp[101][101];
    for(int i=1;i<=N;++i){
        cin >> array[i];
    }
    int l,r,len,k;
    for(len=3;len<=N;++len){
        for(l=1;l+len-1<=N;++l){
            r = l+len-1;
            dp[l][r] = INT_MAX;
            for(k=l+1;k<=r-1;++k){
            dp[l][r] = min(dp[l][r], dp[l][k]+dp[k][r]+array[l]*array[k]*array[r]);
            }
        }
    }
    cout << dp[1][N] << endl;
}