dp[i][j] 表示從第i個人到第j個人的最小花費。。。

僅考慮這j-i+1個人，對第i個人而言，如果第i個人第k個出去，那麼i+1到i+1+k-1-1個人都在第i個人之前出去了，這裡就是一個子問題 dp[i+1][i+1+k-1-1]，i出去要加上(k-1)*D[i]，i+k個人到j個人肯定是在i之後出去的這裡是另一個子問題dp[i+k][j]，不過這些人已經等了k個人所以還要加上sum[i+k..j]×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 
#include 
#include 

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\n",cas++,dp[1][n]);
}
	return 0; 
}