HDU4283 You Are the One（区间dp）

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2714    Accepted Submission(s): 1247

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 

题意：有n个人按1,2,3...n的顺序排好，每个人都有一个unhappy值；如果第i个人第k个上台，那么他的unhappy值为（k-1)*unhappy[i]。他们上台前需要经过一个小黑屋（就相当于是堆栈。。。）

分析：不能用堆栈来写，用了就错了，，，其实是个区间dp，比较难一点；dp[i][j]表示区间[i,j]最小不开心总值。那么对于dp[i][j]的第i个人，就有可能第1个上场，也可以第j-i+1个上场。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][i+1+k-1-1]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[i+k][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(sum[j]-sum[i+k-1])*k。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1<<27;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

int a[105],sum[105],dp[105][105];

int main()
{
    int T,n;
    scanf("%d",&T);
    for(int cas=1; cas<=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];
        }
        CL(dp, 0);
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                dp[i][j] = INF;
        for(int l=1; l<=n-1; l++)
        {
            for(int i=1; i<=n-l; i++)
            {
                int j = i+l;
                for(int k=i; k<=j; k++)
                    dp[i][j] = min(dp[i][j], dp[i+1][k]+dp[k+1][j]+a[i]*(k-i)+(sum[j]-sum[k])*(k-i+1));
            }
        }
        printf("Case #%d: %d\n",cas,dp[1][n]);
    }
    return 0;
}
</span>