POJ1651——Multiplication Puzzle（区间dp）

最新推荐文章于 2023-12-29 13:09:48 发布

原创最新推荐文章于 2023-12-29 13:09:48 发布 · 419 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #HDU #动态规划

算法同时被 3 个专栏收录

246 篇文章

订阅专栏

暑假训练+个人复习

84 篇文章

订阅专栏

动态规划

62 篇文章

订阅专栏

本文介绍了一道经典的区间动态规划题目——乘法谜题。玩家的目标是在一系列正整数卡片中，通过合理的顺序移除卡片以最小化总得分。文章提供了完整的C++代码实现，并详细解释了如何通过动态规划求解最优解。

最近刷区间dp，先写水题

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8604		Accepted: 5375

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

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <iostream>
using namespace std;
const int MAXN =310;
const int INF =1000000007 ;
const int MOD =1000000007;
const double EPS=1e-8;
const double pi = acos(-1);

long long dp[MAXN][MAXN];
long long a[MAXN];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lld",a+i);
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(i>j)
                dp[i][j]=0;
            else
                dp[i][j]=INF;
        }
    }
    for(int i=1;i<n-1;i++)
        dp[i][i]=a[i]*a[i+1]*a[i-1];
    for(int k=1;k<=n-1;k++){
        for(int i=1;i+k<n-1;i++){
            int j=i+k;
            //枚举断点
            for(int m=i;m<=j;m++){
                dp[i][j]=min(dp[i][j],dp[i][m-1]+dp[m+1][j]+a[m]*a[i-1]*a[j+1]);
            }
        }
    }
    printf("%lld\n",dp[1][n-2]);
}