hdu 4283 You Are the One

最新推荐文章于 2021-10-20 16:48:09 发布

threeh20

最新推荐文章于 2021-10-20 16:48:09 发布

阅读量273

点赞数

分类专栏：区间dp

本文链接：https://blog.youkuaiyun.com/threeh20/article/details/77720764

版权

区间dp 专栏收录该内容

9 篇文章

订阅专栏

You Are the One

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4424 Accepted Submission(s): 2074

Problem Description

　　The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?

Input

　　The first line contains a single integer T, the number of test cases. For each case, the first line is n (0 < n <= 100)
　　The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)

Output

　　For each test case, output the least summary of unhappiness .

Sample Input

Sample Output

  
  
   
   Case #1: 20
Case #2: 24

Source

2012 ACM/ICPC Asia Regional Tianjin Online

Recommend

liuyiding | We have carefully selected several similar problems for you: 4267 4268 4269 4270 4271

给出一串数列，给这些数排顺序，代价是排在该数前面的数的数量乘上它本身的数字，求最终的代价和最小
这道题的状态转移方程是dp[i][j]=min(dp[i][j],a[i]*(k-1)+k*(ans[j]-ans[i+k-1])+dp[i+1][i+k-1]+dp[i+k][j])
可以从局部来看，将某一段区间单独来看，假设该段区间的a[i]是在该区间内的第k个，那么i后面的就可以拆成k*(ans[j]-ans[i+k-1])与以i+k到j为
区间的最优解的和。

#include<iostream>
#include<cstring>
using namespace std;
int main(){
    int a[111];
    int dp[111][111];
    int s[111];
    int t,cs=1;
    cin>>t;
    while(t--)
    {
        int num;
        int i,j,k;
        cin>>num;
        s[0]=0;
        for(i=1;i<=num;i++)
        {
        cin>>a[i];
        s[i]=s[i-1]+a[i];
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=num;i++)
        {
            for(j=i+1;j<=num;j++)
            dp[i][j]=99999999;
        }
        for(i=1;i<num;i++)
        {
            for(j=1;j<=num-i;j++)
            {
                int lm=j+i;
                for(k=1;k<=lm-j+1;k++)
                dp[j][lm]=min(dp[j][lm],dp[j+1][j+k-1]+dp[j+k][lm]+k*(s[lm]-s[j+k-1])+a[j]*(k-1));
            }
        }
        cout<<"Case #"<<cs++<<": ";
        cout<<dp[1][num]<<endl;
    }
    return 0;
}