You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3878    Accepted Submission(s): 1793

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

网络赛的题目还真是难理解

题意我就不说了 说下理解吧。

一开始想的时候一直在想怎么维护栈（就是那个黑屋子），然后处理的细节太多不靠谱。这里卡了好久，后来发现思考的方向出了问题

其实每一个男生在上台的时候 最极端的情况有两种 一种是前面的男孩都在小黑屋子里候着，在就是小黑屋子里一个人都没有。其他情况就是在这两种极端情况之间咯，那么这个将要上台男生的上台次序我们就可以枚举确定了， 这一步理解了，就发现是一个典型的区间dp题目了

我们定义dp[i][j]表示序列 i~j的男生上台的最小花费 具体的状态转移看代码~

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int inf=100000009;
int main()
{
    int t;
    cin>>t;
    for(int Case=1;Case<=t;Case++)
    {
        int n;
        cin>>n;
        int a[101],sum[101];
        cin>>a[1];
        sum[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            cin>>a[i];
            sum[i]=sum[i-1]+a[i];
        }
        int dp[120][120];// 注意这里的次序大小都是相对大小 ，不是绝对大小
        memset(dp,0,sizeof(dp));
        for(int l=2;l<=n;l++)
        {
            for(int i=1;i+l-1<=n;i++)
            {
                int j=i+l-1;
                dp[i][j]=inf;
                for(int k=1;k<=l;k++)//枚举i是第几个上场的
                {
                    int temp=i+k;//确定位置
                   /*
                考虑第K个上场即在i+1之后的K-1个人是率先上场的，那么就出现了一个子问题 dp[i+1][temp]表示在第i个人之前上场的
                 对于第i个人，由于是第k个上场的，那么愤怒值便是a[i]*(k-1)
                 其余的人是排在第k+1个之后出场的，也就是一个子问题dp[temp][j]，对于这个区间的人，由于排在第k+1个之后，所以整体愤怒值要加上k*(sum(temp-1))
                    */
                    dp[i][j]=min(dp[i][j],dp[i+1][temp-1]+dp[temp][j]+a[i]*(k-1)+(sum[j]-sum[temp-1])*k);
                }
            }
        }
        printf("Case #%d: %d\n",Case,dp[1][n]);
    }
    return 0;
}

　　dp不能急~ 慢慢啃~