HDOJ 4283 You Are the One

dp[i][j] 表示从第i个人到第j个人的最小花费。。。

仅考虑这j-i+1个人，对第i个人而言，如果第i个人第k个出去，那么i+1到i+1+k-1-1个人都在第i个人之前出去了，这里就是一个子问题 dp[i+1][i+1+k-1-1]，i出去要加上(k-1)*D[i]，i+k个人到j个人肯定是在i之后出去的这里是另一个子问题dp[i+k][j]，不过这些人已经等了k个人所以还要加上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　　512345554322

 

Sample Output

Case #1: 20Case #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\n",cas++,dp[1][n]);}return 0; }