POJ 1651 Multiplication Puzzle(区间DP)

Multiplication Puzzle Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8107 Accepted: 5034 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

3650 区间DP,区间划分寻找最优值就好了

关于区间DP,可以参照这个博客 http://blog.csdn.net/dacc123/article/details/50885903

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>

using namespace std;
#define MAX 10000000
int a[105];
int n;
int dp[105][105];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
       memset(dp,0,sizeof(dp));
        for(int l=2;l<n;l++)
        {
            for(int i=1;i+l<=n;i++)
            {
                int j=i+l;
                dp[i][j]=MAX;
                for(int k=i+1;k<j;k++)
                {
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
                }
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

来自专栏ml

HDUOJ------3336 Count the string(kmp)

D - Count the string Time Limit:1000MS     Memory Limit:32768KB     64bit IO F...

30160
来自专栏用户2442861的专栏

python format 使用 技巧

Simple positional formatting is probably the most common use-case. Use it if the...

45420
来自专栏ml

hdu----(1950)Bridging signals(最长递增子序列 (LIS) )

Bridging signals Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/3...

39370
来自专栏一个会写诗的程序员的博客

Gradle 环境安装Installation

The current Gradle release is 4.10.2. You can download binaries and view docs fo...

20230
来自专栏数据结构与算法

HDU 1506 Largest Rectangle in a Histogram(单调栈)

16900
来自专栏算法修养

POJ-1975 Median Weight Bead(Floyed)

Median Weight Bead Time Limit: 1000MS Memory Limit: 30000K Total Submis...

34980
来自专栏ml

hdu------(1757)A Simple Math Problem(简单矩阵快速幂)

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32...

37370
来自专栏技术小黑屋

Google IO:Android内存管理主题演讲记录

翻出了3年前的Google IO大会的主题演讲 Google IO 2011 Memory management for Android Apps,该演讲介绍...

12910
来自专栏ml

hdu-----(1532)Drainage Ditches(最大流问题)

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/3...

30790
来自专栏算法修养

SPOJ Number of Palindromes(回文树)

Number of Palindromes Time Limit: 100MS Memory Limit: 1572864KB 64bit I...

28560

扫码关注云+社区

领取腾讯云代金券