hdu 6345 Problem J. CSGO

Problem J. CSGO

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 444    Accepted Submission(s): 220

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

 

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思路：

原题可抽象为在多维曼哈顿路径问题下，求最远点距离。

 

在坐标轴中，每个数都有+ -两个位置， 一个k维点则会有 2^k种状态。

我们将问题转化为k+2维曼哈顿路径 ，枚举所有点的 2^(k+2）种状态，求最远点距离。

 

代码如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf=1e18;
const int maxn=1e5+10;
ll a[maxn][7],b[maxn][7],ans,tmp;
ll mavm,mivm,mavs,mivs;
int _,n,m,k;

int main(){
    for (scanf("%d",&_); _; _--){
        ans=-inf;
        scanf("%d%d%d",&n,&m,&k);
        for (int i=1; i<=n; i++){
            scanf("%lld",&a[i][0]);
            for (int j=2; j<=k+1; j++)
                scanf("%lld",&a[i][j]);
        }
        for (int i=1; i<=m; i++){
            scanf("%lld",&b[i][1]);
            for (int j=2; j<=k+1; j++)
                scanf("%lld",&b[i][j]);
        }
        int upp=(1<<k+2);
        for (int d=0; d<upp; d++){
            mavm=-inf,mavs=-inf;
            mivm=inf,mivs=inf;
            for (int i=1; i<=n; i++){
                tmp=0;
                for (int j=0; j<k+2; j++){
                    if(d&(1<<j)) tmp+=a[i][j];
                    else tmp-=a[i][j];
                }
                mavm=max(mavm,tmp);
                mivm=min(mivm,tmp);
            }
            for (int i=1; i<=m; i++){
                tmp=0;
                for (int j=0; j<k+2; j++){
                    if(d&(1<<j)) tmp+=b[i][j];
                    else tmp-=b[i][j];
                }
                mavs=max(mavs,tmp);
                mivs=min(mivs,tmp);
            }
            ans=max(ans,max(mavm-mivs,mavs-mivm));
        }
        printf("%lld
",ans);
    }
    return 0;
}