hdu 4283 You Are the One(区间DP)题目转换难，状态难，。。。

1、http://acm.hdu.edu.cn/showproblem.php?pid=4283

2、题目大意：

有n个男孩，现在要一个一个的登台，每个男孩都有一个diaosi值，如果他是第k个登台的，他的最终的diaosi值就等于他的diaosi值*（k-1),他前边等待了K-1个人，现在要确定一个登台顺序 ，使得所有男孩的diaosi值最小

3、思路分析

设状态dp[i][j]表示i-j区间内i是第k个登台的

 dp[s][e]=min(dp[s][e],dp[s+1][e]+a[s]*(e-s));//s是最后一个选出来的
            dp[s][e]=min(dp[s][e],dp[s+1][e]+Delay(s,e));//第一个被选出来
            for(int k=s+1;k<=e;k++)
            {
                dp[s][e]=min(dp[s][e],dp[s+1][k]+dp[k+1][e]+a[s]*(k-s)+Delay(k,e)*(k-s+1));
                //最后是k-s+1,k后边的人还要等k
            }

4、题目：

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1351    Accepted Submission(s): 637

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

 

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5、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
#define INF 1<<29
int a[N],sum[N];
int dp[N][N];//dp[i][j]表示i是第k个被选出来的
int Delay(int s,int e)
{
    if(s>e)
    return 0;
    return sum[e]-sum[s];
}
void DP(int n)
{
    for(int i=0;i<=n;i++)
    {
        for(int  j=0;j<=n;j++)
        dp[i][j]=INF;
    }
    for(int i=0;i<=n;i++)
    {
        dp[i][i]=0;
    }
    for(int d=1;d<=n;d++)
    {
        for(int i=1;i+d-1<=n;i++)
        {
            int s=i;
            int e=i+d-1;
            //int delay=Delay(s,e);
            dp[s][e]=min(dp[s][e],dp[s+1][e]+a[s]*(e-s));//s是最后一个选出来的
            dp[s][e]=min(dp[s][e],dp[s+1][e]+Delay(s,e));//第一个被选出来
            for(int k=s+1;k<=e;k++)
            {
                dp[s][e]=min(dp[s][e],dp[s+1][k]+dp[k+1][e]+a[s]*(k-s)+Delay(k,e)*(k-s+1));
                //最后是k-s+1,k后边的人还要等k
            }
        }
    }
}
int main()
{
    int t,n,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        cas++;
        scanf("%d",&n);
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        DP(n);
        printf("Case #%d: %d\n",cas,dp[1][n]);
    }
    return 0;
}
/*
20
5
1 2 3 4 5
5
5 4 3 2 2
6
1 1 1 1 1 1
7
7 7 7 7 7 7 7
*/