hdu4283 区间dp

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

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

/**hdu4283 区间dp题目大意：一些屌丝排队进场，第k个进场的人后又k-1*a[i]的愤怒值，为了得到最小的愤怒值，可以利用一个栈来调整顺序，第i个人进栈可以让第i+1个人先          行入场，对于栈里的元素必须是后进先出，问如何合理利用栈来以得到最小的愤怒值解题思路：我们用dp[i][j]表示区间i~j之中的元素可得到的最小的愤怒值。对于i~j中的元素i我们然他第k个入场，那么其后面的k-1个元素就要先行入场，          这是问题就变成了dp[i+1][i+k-1]和dp[i+k][j], 对于第i个元素的愤怒值为：（k-1）*a[i],而第i+k~j的愤怒值要加上k*(sum[j]-sum[i+1-1]);          最后dp[1][n]就是答案*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;const int maxn=105;int n,a[maxn],sum[maxn],dp[105][105];int main(){    int T,tt=0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        memset(sum,0,sizeof(sum));        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            sum[i]=sum[i-1]+a[i];        }        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)                dp[i][j]=0x3f3f3f;        }        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]+(k-i+1)*(sum[j]-sum[k])+(k-i)*a[i]);                }            }        }        printf("Case #%d: %d\n",++tt,dp[1][n]);    }    return 0;}/**2512345554322*/