hdu-6435

Problem J. CSGO

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 272    Accepted Submission(s): 135

Problem Description

You are playing CSGO.
There are n Main Weapons and m Secondary Weapons in CSGO. You can only choose one Main Weapon and one Secondary Weapon. For each weapon, it has a composite score S.
The higher the composite score of the weapon is, the better for you.
Also each weapon has K performance evaluations x[1], x[2], …, x[K].(range, firing rate, recoil, weight…)
So you shold consider the cooperation of your weapons, you want two weapons that have big difference in each performance, for example, AWP + CZ75 is a good choose, and so do AK47 + Desert Eagle.
All in all, you will evaluate your weapons by this formula.(MW for Main Weapon and SW for Secondary Weapon)

Now you have to choose your best Main Weapon & Secondary Weapon and output the maximum evaluation.

 

Input

Multiple query.
On the first line, there is a positive integer T, which describe the number of data. Next there are T groups of data.
for each group, the first line have three positive integers n, m, K.
then, the next n line will describe n Main Weapons, K+1 integers each line S, x[1], x[2], …, x[K]
then, the next m line will describe m Secondary Weapons, K+1 integers each line S, x[1], x[2], …, x[K]
There is a blank line before each groups of data.
T<=100, n<=100000, m<=100000, K<=5, 0<=S<=1e9, |x[i]|<=1e9, sum of (n+m)<=300000

 

Output

Your output should include T lines, for each line, output the maximum evaluation for the corresponding datum.

 

Sample Input

2
2 2 1
0 233
0 666
0 123
0 456
2 2 1
100 0 1000 100 1000 100
100 0

 

Sample Output

543
2000

 

Source

2018 Multi-University Training Contest 10。

 

 

 

　　　　由于| a[i]-b[i] | = max(a[i]-b[i],b[i]-a[i]) ，也就是说主武器和副武器的各个属性前面的符号是相反的，属性数量很少，可以枚举出所有的情况选出一个最优的就是答案。

　　　　

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define mp make_pair
 #define pb push_back
 #define inf 0x7fffffffff
 #define pii pair<int,int>
 int x[][];
 LL a[]={};
 int main()
 {
     int t,n,m,i,j,k;
     cin>>t;
     while(t--){
         scanf("%d%d%d",&n,&m,&k);
         for(i=;i<=n;++i)
             for(j=;j<k+;++j) scanf("%d",&x[i][j]);
         for(i=;i<=m;++i)
             for(j=;j<k+;++j) scanf("%d",&x[i+n][j]);
         LL ans=-inf;
         for(i=;i<(<<k);++i){
             for(j=;j<k;++j)a[j+]=(i&(<<j))?:-;
             LL mx1=-inf,mx2=-inf,tmp=;
             for(j=;j<=n;++j){
                 tmp=x[j][];
                 for(int o=;o<k+;++o){
                     tmp+=a[o]*x[j][o];
                 }
                 if(tmp>mx1)mx1=tmp;
             }
             for(j=n+;j<=n+m;++j){
                 tmp=x[j][];
                 for(int o=;o<k+;++o){
                     tmp-=a[o]*x[j][o];
                 }
                 if(tmp>mx2)mx2=tmp;
             }
             if(mx1+mx2>ans)ans=mx1+mx2;
         }
         cout<<ans<<endl;
     }
     return ;
 }