[Yzoj1579] & [bzoj2450] arr

Description

A sequence of 3 * n numbers Xi is given, which requires that three sequences of AI, Bi, and Ci with N length be constructed to meet the following conditions:

Each number of 1 to 3 * n appears once and only once in the three sequences;

S = sum (X [ai]-X [bi]) * X [CI]) is the largest.

The maximum output value is S. Multiple groups of data.

Input Format

The first line contains two numbers T and N, and T is the number of data groups. N is described as a question.

In the next T line, each line contains 3 * n numbers, indicating Xi.

Output Format

The output contains t rows, with the maximum output s for each row.

Sample Input

1 24 1 8 2 0 5

Sample output

46

Hint

No more than 1000 groups of data for 1 <= n <= 10;

For 11 <= n <= 15, no more than 100 groups of data;

For 16 <= n <= 20, there are no more than 10 groups of data;

For 21 <= n <= 25, there is only one group of data.

All xi <= 1000.

 

Solution

First, orz yff, how to think of DFS... The standard solution is like pressing DP or getting stuck, and I do not know where it is slower than the brute force.

1. When a> = C> = B (a-B) * C is the largest. Just push it out.

2. Sort the number of 3 * n read records in ascending order, X1 ~ XN must be in each group, and it must be B of each group.

3. Sort n tuples (AI, Bi, and CI) in ascending order of Ci. A1 <= a2 <= A3 <=... <= An, B1> = b2 >=…> = Bn. So b1 = xn, b2 = xn-1... BN = x1.

4. Optimum pruning: Two tuples (AI, Bi, CI) and (AJ, BJ, CJ) are the final answers.

(AI-Bi) * CI + (AJ-BJ) * CJ> (AI-Bi) * AJ + (CI-BJ) * CJ

That is, (CI-AJ) (AI-bi-CJ)> 0

Then let's talk about DFS =

 

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 int T,n,a[30],c[30],t[100],ans,now;bool vis[100]; 5 bool check(int x) 6 { 7     for(int i=1;i<x;i++) 8     { 9         if(c[i]>a[x]&&a[i]-t[3*n-i+1]<c[x])return 0;10         if(c[i]<a[x]&&a[i]-t[3*n-i+1]>c[x])return 0;11     }12     return 1;13 }14 void dfs(int xa,int tc)15 {16     if(xa==n+1)17     {18         if(ans<now)ans=now;19         return;20     }21     for(int i=xa;i<=2*n;i++)22     if(!vis[i])23     {24         vis[i]=1,a[xa]=t[i];25         for(int j=std::max(i,tc)+1;j<=2*n;j++)26         if(!vis[j])27         {28             vis[j]=1,c[xa]=t[j],29             now+=(a[xa]-t[3*n-xa+1])*c[xa];30             if(now*n>ans*xa&&check(xa))dfs(xa+1,j);31             now-=(a[xa]-t[3*n-xa+1])*c[xa],vis[j]=0;32         }33         vis[i]=0;break;34     }35 }36 bool cmp(int i,int j){return i>j;}37 int main()38 {39     for(scanf("%d%d",&T,&n);T;T--)40     {41         for(int i=1;i<=3*n;i++)scanf("%d",&t[i]);42         std::sort(t+1,t+1+3*n,cmp);ans=0;43         dfs(1,0);printf("%d\n",ans);44     }45 }

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