CSGO

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

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

题意：求上面那个表达式的最大值。

这个表达式的后面很像曼哈顿距离，所以可以转换为求最远曼哈顿距离。

如何处理前面那两个正数：在每个点上再加一维，主武器权值为正，副武器权值为负。

如何让程序选的两个点分别在主武器集合和副武器集合里：更改主武器的某一维的权值，让其加上一个INF，这样选的最远曼哈顿距离肯定不会是同一个集合里的了，最后答案减去INF。

#include <stdio.h>
#define INF 100000000000
#define M 200005//坐标个数

int D = 6;//空间维数
struct Point
{
    long long x[6];
}pt[M];

long long dis[1<<6][M],minx[1<<6],maxx[1<<6];//去掉绝对值后有2^D种可能
void Get(int N)//取得所有点在指定状态（S）下的“本点有效距离”
{
    int tot=(1<<D);
    for(int s=0;s<tot;s++)//遍历所有维数正负状态
    {
        int coe[D];
        for(int i=0;i<D;i++)//设定当前状态的具体正负参数
        if(s&(1<<i)) coe[i]=-1.0;
        else coe[i]=1.0;

        for(int i=0;i<N;i++)//遍历所有点，确定每个点在当前状态下的“本点有效距离”
        {
            dis[s][i]=0;
            for(int j=0;j<D;j++)
            dis[s][i]=dis[s][i]+coe[j]*pt[i].x[j];
        }
    }
}
//取每种可能中的最大差距
void Solve(int N)
{
    int tot=(1<<D);
    long long tmp,ans;
    for(int s=0;s<tot;s++)
    {
        minx[s]=INF;
        maxx[s]=-INF;
        for(int i=0;i<N;i++)
        {
            if (minx[s]>dis[s][i]) minx[s]=dis[s][i];
            if (maxx[s]<dis[s][i]) maxx[s]=dis[s][i];
        }
    }
    ans=0;
    for(int s=0;s<tot;s++)
    {
        tmp=maxx[s]-minx[s];
        if(tmp>ans) ans=tmp;
    }

    printf("%lld
", ans-INF);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,k;
        scanf("%d %d %d",&n,&m,&k);
        D=k+1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=k;j++)
            {
                scanf("%lld",&pt[i].x[j]);
            }
            pt[i].x[0]+=INF;
        }

        for(int i=0;i<m;i++)
        {
            for(int j=0;j<=k;j++)
            {
                scanf("%lld",&pt[i+n].x[j]);
            }
            pt[i+n].x[0]=-pt[i+n].x[0];
        }
        Get(n+m);
        Solve(n+m);
    }
    return 0;
}

View Code