【hdu 6435】杭电多校：CSGO（曼哈顿距离+二进制枚举）

题目描述

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.

 

输入

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

 

输出

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

 

样例输入

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

 

样例输出

543
2000

题意：

　　就是给你一堆主武器和副武器，然后挑选其中综合得分最高的（主副武器本身分值高，且差值最大），计算公式如上。

思路：

　　既然要挑选武器，最好的方法就是最大的减去最小的。所以，可以这么考虑，对于每一种属性，它对于答案的贡献，要么是加，要么是减

　　所以可以通过二进制枚举，枚举每一种状态，对于主武器的s，和副武器的s，将他们考虑进来，将他们的位置错开保证他们不会相减,即

　　主s设在0位置，副武器s设在1位置，所以共k+2个属性，也就是2^(k+2)种枚举。

代码：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+50;
typedef long long ll;
int n,m,k,top;
ll ma[maxn][10],se[maxn][10];
ll maxv1,minv1,maxv2,minv2;
ll all;
void read1()
{
    for(int i=0; i<n; i++)
    {
        scanf("%lld",&ma[i][0]);
        for(int j=0; j<k; j++)
        {
            scanf("%lld",&ma[i][j+2]);
        }
    }
}
void read2()
{
    for(int i=0; i<m; i++)
    {
        scanf("%lld",&se[i][1]);
        for(int j=0; j<k; j++)
        {
            scanf("%lld",&se[i][j+2]);
        }
    }
}
void solve1()
{
    for(int q=0; q<top; q++)
    {
        maxv1=-1e18,minv1=1e18;
        maxv2=-1e18,minv2=1e18;
        for(int i=0; i<n; i++)
        {
            ll temp=0;
            for(int j=0; j<k+2; j++)
            {
                if(q&(1<<j))
                {
                    temp+=ma[i][j];
                }
                else
                {
                    temp-=ma[i][j];
                }
            }
            maxv1=max(maxv1,temp);
            minv1=min(minv1,temp);
        }
        for(int i=0; i<m; i++)
        {
            ll temp=0;
            for(int j=0; j<k+2; j++)
            {
                if(q&(1<<j))
                {
                    temp+=se[i][j];
                }
                else
                {
                    temp-=se[i][j];
                }
            }
            maxv2=max(maxv2,temp);
            minv2=min(minv2,temp);
        }
        ll maxx=max(maxv1-minv2,maxv2-minv1);
        all=max(all,maxx);
    }
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        read1();
        read2();
        all=0;
        top=1<<(k+2);
        solve1();
        //solve2();
        cout << all << endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}
/*
2
2 2 1
0 233
0 666
0 123
0 456
2 2 1
100 0 1000 100 1000 100
100 0
*/
/*
546
2000
*/