dp[i][j]는 i개인에서 j개인까지의 최소 비용을 나타낸다.
이 j-i+1명만 생각해서 i개인에게 만약에 i개인이 k번째로 나가면 i+1부터 i+1+k-1-1까지 모두 i개인이 나가기 전에 나간다. 여기는 바로 하나의 문제인 dp[i+1][i+1+k-1]이다. i가 나가면 (k-1)*D[i]를 더해야 한다. i+k가 j개인이 i에 들어가면 i가 나간 것이 틀림없다. 여기는 또 다른 문제인 dp[i+k][j]이다. 그러나 이 사람들은 이미 k사람을 기다렸기 때문에sum[i+k]를 추가해야 한다.×k

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1381    Accepted Submission(s): 652
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 <cstdio>
#include <cstring>

using namespace std;

const int INF=0x3f3f3f3f;

int n,D[120],sum[120];
int dp[120][120];

int main()
{	
	int T_T,cas=1;
	scanf("%d",&T_T);
while(T_T--)
{
	scanf("%d",&n);
	sum[0]=0;
	memset(dp,0,sizeof(dp));
	for(int i=1;i<=n;i++) 
	{
		scanf("%d",D+i);
		sum[i]=sum[i-1]+D[i];
	}
	for(int i=0;i<120;i++) for(int j=i+1;j<120;j++) dp[i][j]=INF;
	for(int len=2;len<=n;len++)
	{
 		for(int i=1;i+len-1<=n;i++)
 		{
 			int j=i+len-1;
 			for(int k=1;k<=len;k++)
 			{
 				dp[i][j]=min(dp[i][j],dp[i+1][i+k-1]+(k-1)*D[i]+dp[i+k][j]+k*(sum[j]-sum[i+k-1]));
 			}
 		}
	}
	printf("Case #%d: %d
",cas++,dp[1][n]);
}
	return 0; 
}