You Are the One（利用栈的性质的区间DP）

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

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2436    Accepted Submission(s): 1137

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

 

以下分析来自： http://blog.csdn.net/u014141559/article/details/38035603?utm_source=tuicool

题目大意：

有n个男屌丝事先按1,2,3,,,,,,n的顺序排好，每个人都有一个不开心值unhappy[i]，如果第i个人第k个上台找对象，那么该屌丝男的不开心值就会为（k-1)*unhappy[i]，因为在他前面有k-1个人嘛，导演为了让所有男屌的总不开心值最小，搞了一个小黑屋，可以通过小黑屋来改变男屌的出场顺序

注意：这个小黑屋是个栈，男屌的顺序是排好了的，但是可以通过入栈出栈来改变男屌的出场顺序

解题思路：（操度娘所知~度娘你好腻害）

dp[i][j]表示区间[i,j]的最小总不开心值

把区间[i,j]单独来看，则第i个人可以是第一个出场，也可以是最后一个出场（j-i+1），也可以是在中间出场（1   ~  j-i+1）

不妨设他是第k个出场的（1<=k<=j-i+1)，那么根据栈后进先出的特点，以及题目要求原先男的是排好序的，那么：：

第  i+1  到 i+k-1  总共有k-1个人要比i先出栈，

第 i+k   到j 总共j-i-k+1个人在i后面出栈

举个例子吧：

有5个人事先排好顺序  1,2,3,4,5

入栈的时候，1入完2入，2入完3入，如果我要第1个人第3个出场，那么入栈出栈顺序是这样的：

1入，2入，3入，3出，2出，1出（到此第一个人就是第3个出场啦，很明显第2，3号人要在1先出，而4,5要在1后出）

这样子，动态转移方程就出来啦，根据第i个人是第k个出场的，将区间[i,j]分成3个部分

dp[i][j]=min(dp[i][j],dp[i+1,i+k-1]+dp[i+k,j]+(k-1)*a[i]+(sum[j]-sum[i+k-1])*k);   

(sum[j]-sum[i+k-1])*k 表示 后面的 j-i-k+1个人是在i后面才出场的，那么每个人的不开心值都会加个 unhappy，sum[i]用来记录前面i个人的总不开心值，根据题目，每个人的unhappy是个累加的过程，多等一个人，就多累加一次

AC code：

#include
#include
#include
#include
#include
#include
#include
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
int dp[111][111];
int sum[111];
int a[111];
int n;
int main()
{
    int T,cas,i,j,k,p;
    cas=0;
    scanf("%d",&T);
    while(T--)
    {
        cas++;
        sum[0]=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                dp[i][j]=INF;
            }
        }
        for(int p=1;p<=n;p++)
        {
            for(i=1;i<=n-p+1;i++)
            {
                j=i+p-1;
                for(k=1;k<=p;k++)
                {
                    dp[i][j]=min(dp[i][j],dp[i+1][i+k-1]+dp[i+k][j]+(k-1)*a[i]+(sum[j]-sum[i+k-1])*k);

                }
            }
        }
        printf("Case #%d: %d\n",cas,dp[1][n]);

    }
    return 0;
}

记忆化搜索的区间DP比较好理解，附上：

#include
#include
#include
#include
using namespace std;
int dp[110][110];
int a[110],sum[110];
int n;
int solve(int i,int j)
{
    int &ans=dp[i][j];
    if(ans!=-1) return ans;
    if(i>=j) return 0;
    ans=1<<30;
    for(int k=1;k<=j-i+1;k++){
        ans=min(ans,solve(i+1,i+k-1)+solve(i+k,j)+(k-1)*a[i]+(sum[j]-sum[i+k-1])*k);
    }
    return ans;
}
int main()
{
    int t,iCase=1;
    cin>>t;
    while(t--){
        cin>>n;
        sum[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        memset(dp,-1,sizeof dp);
        printf("Case #%d: %d\n",iCase++,solve(1,n));
    }
    return 0;
}