hdu 4283 动态规划 类似括号匹配

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4283

题意：

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 450    Accepted Submission(s): 246

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

   
   
   
   

    
    
    
    2
　　
5
1
2
3
4
5

5
5
4
3
2
2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 20
Case #2: 24
   
   
   
   

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int dp[102][102],t,n,num[102],sum[102],cas=1;
// dp[i][j] 表示i  在区间[i,j]中 是第k个选出来的！！
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        int i,j,k,tmp;
        sum[0]=0;
        for(i=1; i<=n; i++)
        {
            cin>>num[i];
            sum[i]=sum[i-1]+num[i];
        }
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                if(i==j) dp[i][j]=0;
                else dp[i][j]=(1<<29);
            }
        }
        for(k=1; k<n; k++)
        {
            for(i=1; i<=n-k; i++)
            {
                int r=i+k;
                dp[i][r]=min(dp[i][r],dp[i+1][r]+num[i]*k);// 区间里的最后一个
                dp[i][r]=min(dp[i][r],dp[i+1][r]+sum[r]-sum[i]);//区间里的第一个
                for(j=i+1; j<=r; j++)
                {
                    dp[i][r]=min(dp[i][r],dp[i+1][j]+dp[j+1][r]+num[i]*(j-i)+(sum[r]-sum[j])*(j-i+1));
                }
            }
        }
        printf("Case #%d: %d\n",cas++,dp[1][n]);
    }
    return 0;
}
/*
5　　
5
1
2
3
4
5

5
5
4
3
2
2
*/