D - Multiplication Puzzle

最新推荐文章于 2020-10-22 16:26:20 发布

JustSmilelee

最新推荐文章于 2020-10-22 16:26:20 发布

阅读量675

点赞数

CC 4.0 BY-SA版权

分类专栏：区间dp

本文链接：https://blog.youkuaiyun.com/Limhhhhh/article/details/50600725

区间dp 专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种使用区间动态规划方法解决特定乘法谜题的算法实现。该问题要求玩家从一排含有正整数的卡片中，按照一定顺序抽取卡片以使总得分最小。文章提供了完整的C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

D - Multiplication Puzzle

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

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

很多人都说这道题是矩阵连乘，但是我不是很清楚矩阵连乘是怎样的方法，所以就还是当做区间dp把。题意是要把除了两端的剩下的牌一张一张的抽出，每抽出一张就要加上当前牌和他的左右两张牌共3张牌的乘积，要求最后得到的总数最小是多少。

</pre><pre name="code" class="cpp">#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define inf 1e8
int main()
{
    int dp[105][105],a[105],n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int l=3;l<=n;l++)      //需要去掉左右端点，最短长度就是3
        {
            for(int i=0;i+l-1<n;i++)
            {
                int j=i+l-1;
                dp[i][j]=inf;
                for(int k=i+1;k<j;k++)      //枚举中间的位置k
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}