You Are the One - HDU 4283 dp

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1714    Accepted Submission(s): 818

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　　512345554322

 

Sample Output

Case #1: 20Case #2: 24

 

题意：每个人的不满意度是前面的人数*Di，可以通过一个栈去给他们排序，问最小的不满意度是多少。

思路：区间dp，对于一个区间，要么是把第一个放到最后，要么是从中间分开。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int T,t,n,dp[110][110],val[110],sum[110];int main(){    int i,j,k,len;    scanf("%d",&T);    for(t=1;t<=T;t++)    {        scanf("%d",&n);        for(i=1;i<=n;i++)        {            scanf("%d",&val[i]);            sum[i]=sum[i-1]+val[i];        }        for(len=1;len<n;len++)           for(i=1;i+len<=n;i++)           {               j=i+len;               dp[i][j]=dp[i+1][j]+(j-i)*val[i];               for(k=i;k<j;k++)                  dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+(k-i+1)*(sum[j]-sum[k]));           }        printf("Case #%d: %d\n",t,dp[1][n]);    }}