Problem J. CSGO

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 228 Accepted Submission(s): 109

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 1

0 233

0 666

0 123

0 456

2 2 1

100 0

1000 100

100 0

Sample Output

543

2000

题意：

有n件主武器和m件副武器，所有武器都会有一个价值S和K种属性,当你选择一件主武器i和一件副武器j,得到的价值为 $S_{i}+S_{j}+\sum _{i=1}^{K}|x_{i}-x_{j}|$ ,也就是两个武器的S要大,两件武器属性差要大。

做法：

最远n维曼哈顿距离。

用tmp[x]表示所有主武器的K种属性在状态为x(转为二进制)时的最大贡献。

比如数2 3 4 在状态001下就是(-2-3+4),0表示减，1表示加。

在得到(1<<k)个tmp后,对于副武器中的一个i的K种属性,也枚举 $2^{k}$ 个状态,然后和之前主武器中的相反状态相加，与之前的状态取一个异或值，维护答案取一个最大值即可。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,k,save[35],tmp[10],ans;
int main(){
    int t;
    cin>>t;
    while(t--){
        scanf("%lld%lld%lld",&n,&m,&k);
        memset(save,-125,sizeof(save));
        for(int j=1;j<=n;j++){
            scanf("%lld",&ans);
            for(int i=1;i<=k;i++)
                scanf("%lld",&tmp[i]);
            for(int i=0;i<(1<<k);i++){
                ll tmpans=ans;
                for(int z=1;z<=k;z++){
                    if(i&(1<<(z-1))) tmpans+=tmp[z];
                    else tmpans-=tmp[z];
                }
                save[i]=max(save[i],tmpans);
            }
        }
        ll aans=0;
        for(int j=1;j<=m;j++){
            scanf("%lld",&ans);
            for(int i=1;i<=k;i++)
                scanf("%lld",&tmp[i]);
            for(int i=0;i<(1<<k);i++){
                ll tmpans=ans;
                for(int z=1;z<=k;z++){
                    if(i&(1<<(z-1))) tmpans-=tmp[z];
                    else tmpans+=tmp[z];
                }
               aans=max(aans,save[i]+tmpans);
            }
        }
        printf("%lld\n",aans);
    }
    return 0;
}

hdu 6435-Problem J. CSGO N维曼哈顿最远距离+思维

Problem J. CSGO

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