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?

 

　　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)

 

　　For each test case, output the least summary of unhappiness .

 

 

 

2012 ACM/ICPC Asia Regional Tianjin Online

 

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题意：有n个人，输入n个基值（愤怒值）v[]，表示每个人的愤怒值，现在他们依次上台表演，如果i是第k个上台，那么他的愤怒值为（k-1）*v[i]  现在有一个栈，可以让一些人进栈，让后面的人先上台表演，这样可以改变部分顺序，求所有人最小的愤怒值和；

 

思路：区间DP，定义dp[i][j]表示原序列中i到j的人的最小愤怒值的和，设i在区间i到j中第k个上场，那么i+1~i+k-1会先i上台，然后i上台，最后i+k~i+len上台，那么状态转移方程为： dp[i][j]=v[i]*(k-1)+dp[i+1][i+k-1]+(sum[i+len]-sum[i+k-1])*k+dp[i+k][i+len]  sum[]表示前缀和;

 

代码如下：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <set>using namespace std;const int inf=0x7fffffff;int v[105];int dp[105][105];int sum[105];int main(){    int T,n,Case=1;    cin>>T;    while(T--)    {        scanf("%d",&n);        sum[0]=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&v[i]);            sum[i]=sum[i-1]+v[i];        }        for(int i=1;i<=n;i++)            dp[i][i]=0;        for(int len=1;len<n;len++)        {            for(int i=1;i+len<=n;i++)            {   ///特判i在i~i+len的区间中是第一个和最后一个时的情况；                dp[i][i+len]=min(sum[i+len]-sum[i]+dp[i+1][i+len],dp[i+1][i+len]+v[i]*len);                for(int k=2;k<=len;k++)                dp[i][i+len]=min(dp[i][i+len],v[i]*(k-1)+dp[i+1][i+k-1]+(sum[i+len]-sum[i+k-1])*k+dp[i+k][i+len]);            }        }        printf("Case #%d: %d\n",Case++,dp[1][n]);    }}