参考：https://blog.csdn.net/ffgcc/article/details/81951339
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.
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
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：
给你n件主武器和m件副武器 每件武器都有1个综合评定分数S 和k个其他参数x[i]
你要选择一件主武器和一件副武器 两件武器搭配的值为(MW为主武器 SW为副武器)
分析：
最终的答案的选择中 主副武器的x[i]总是一个加对应一个减 一个减对应一个加
所以只需要枚举出2^k种情况 选最大的就可以了
代码：
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5+5; const int inf = 0x3f3f3f3f; typedef long long ll; ll s1[maxn]; ll s2[maxn]; ll v1[maxn][10]; ll v2[maxn][10]; ll n,m,k; int T; int main() { scanf("%d",&T); while(T--) { scanf("%lld%lld%lld",&n,&m,&k); for(ll i=0;i<n;i++) { scanf("%lld",&s1[i]); for(ll j=0;j<k;j++) { scanf("%lld",&v1[i][j]); } } for(ll i=0;i<m;i++) { scanf("%lld",&s2[i]); for(ll j=0;j<k;j++) { scanf("%lld",&v2[i][j]); } } ll s = 1<<k; //(共有2的k次情况 每种都有一个最大值 最后选最大的) ll maxans = -inf; ll tmp; for(ll i=0;i<s;i++) { ll max1 = -inf; ll max2 = -inf; for(ll j=0;j<n;j++) { tmp = s1[j]; for(ll q=0;q<k;q++) { if((1<<q)&i) tmp+=v1[j][q]; //v1 v2加减相反 else tmp-=v1[j][q]; } max1 = max(max1,tmp); } for(ll j=0;j<m;j++) { tmp = s2[j]; for(ll q=0;q<k;q++) { if((1<<q)&i) tmp-=v2[j][q]; else tmp+=v2[j][q]; } max2 = max(max2,tmp); } maxans = max(maxans,max1+max2); } printf("%lld ",maxans); } return 0; }
注：菜鸡的我看了一上午这道题 一直不会这种按位枚举的方法
对于这道题 举个栗子吧
若n = 5 v1[][]的加减情况与i的关系（v2[][]相反）
i：0~ 2^5-1 即可枚举出所有情况