hdu: You Are the One（区间DP）

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 .

输入样例

2
　　
5
1
2
3
4
5

5
5
4
3
2
2
 

输出样例

Case #1: 20
Case #2: 24

子区间划分：由于栈的特点，区间的第一个元素在第k位出栈，则（1，k-1）（k+1,n）互相独立
注意dp数组的赋值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=400,inf=0x3f3f3f3f;
int a[N],s[N],dp[N][N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int T;cin>>T;
    for(int t=1;t<=T;++t)
    {
        int n;
        cin>>n;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;++i)
        {cin>>a[i];
        s[i]=a[i]+s[i-1];
        }
        for(int i = 0; i <= n; i++)
         for(int j = i+1; j <= n; j++)
                dp[i][j] = inf;//注意初始化只有成立区间赋值为inf
        //dp[i][j]表示第i个人以i为起点
        //第k个出场，k从i到j的相对焦虑最小值
        for(int len=2;len<=n;++len)
         for(int i=1;i<=n;++i)
        {
            int j=i+len-1;
            if(j>n) break;
            for(int k=i;k<=j;++k)//第i个人在第k个出场
            {
                dp[i][j]=min(dp[i][j],(k-i)*a[i]+dp[i+1][k]+dp[k+1][j]
                             +(k-i+1)*(s[j]-s[k]));
                //i在第k个出场，则子区间为(i+1,k) (k+1,j)
            }
        }
        cout<<"Case #"<<t<<": "<<dp[1][n]<<'\n';
    }
    return 0;
}