You Are the One - HDU 4283 dp

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本文探讨了一个在排队系统中通过调整队列顺序来最小化个体等待不满意度的问题，具体涉及到队列中个体的个性差异及其对整体等待体验的影响。

You Are the One

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1829 Accepted Submission(s): 866

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

Sample Output

Case #1: 20
Case #2: 24

题意：在一个队列中，有n个人，每个人每等一个人就会增加相应的不满意度，你有一个栈可以改变队列的顺序，求最小的不满意度。

思路：dp[i][j]表示如果只有i-j的人的话，不满意度最小为多少，每次转移的时候，枚举第i个人放在第k个位置，然后算出相应的不满意度即可。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int T,t,n,num[110],sum[110],dp[110][110],INF=1e9;
int main()
{
    int i,j,k,d;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            sum[i]=sum[i-1]+num[i];
        }
        for(d=1;d<=n;d++)
           for(i=1;i+d<=n;i++)
           {
               j=i+d;
               dp[i][j]=INF;
               for(k=i;k<=j;k++)
                  dp[i][j]=min(dp[i][j],dp[i+1][k]+(k-i)*num[i]+dp[k+1][j]+(sum[j]-sum[k])*(k-i+1));
           }
        printf("Case #%d: %d\n",t,dp[1][n]);
    }
}