题目描述
Winter holidays are coming up. They are going to last for nn days.
During the holidays, Monocarp wants to try all of these activities exactly once with his friends:
- go skiing;
- watch a movie in a cinema;
- play board games.
Monocarp knows that, on the i -th day, exactly ai friends will join him for skiing, bi friends will join him for a movie and cici friends will join him for board games.
Monocarp also knows that he can't try more than one activity in a single day.
Thus, he asks you to help him choose three distinct days x,y,z in such a way that the total number of friends to join him for the activities ( ax+by+cz ) is maximized.
输入格式
The first line contains a single integer t ( 1≤t≤104 ) — the number of testcases.
The first line of each testcase contains a single integer n ( 3≤n≤105 ) — the duration of the winter holidays in days.
The second line contains nn integers a1,a2,…,an ( 1≤ai≤108 ) — the number of friends that will join Monocarp for skiing on the i -th day.
The third line contains nn integers b1,b2,…,bn ( 1≤bi≤108 ) — the number of friends that will join Monocarp for a movie on the i -th day.
The fourth line contains nn integers c1,c2,…,cn ( 1≤ci≤108 ) — the number of friends that will join Monocarp for board games on the i -th day.
The sum of n over all testcases doesn't exceed 105 .
输出格式
For each testcase, print a single integer — the maximum total number of friends that can join Monocarp for the activities on three distinct days.
题意翻译
给定三个数组 a,b,c，找三个互不相同的整数 i,j,k，使得 ai+bj+ck 的值最大。
输入输出样例
输入 #1复制
4 3 1 10 1 10 1 1 1 1 10 4 30 20 10 1 30 5 15 20 30 25 10 10 10 5 19 12 3 18 18 6 17 10 13 15 17 19 11 16 3 11 17 17 17 1 17 18 10 15 8 17 3 13 12 10 17 5 4 18 12 4 11 2 16 16 8 4 14 19 3 12 6 7 5 16 3 4 8 11 10 8 10 2 20 3
输出 #1复制
30 75 55 56
说明/提示
In the first testcase, Monocarp can choose day 2 for skiing, day 1 for a movie and day 3 for board games. This way, a2=10 friends will join him for skiing, b1=10 friends will join him for a movie and c3=10 friends will join him for board games. The total number of friends is 30.
In the second testcase, Monocarp can choose day 1 for skiing, day 4 for a movie and day 2 for board games. 30+20+25=75 friends in total. Note that Monocarp can't choose day 1 for all activities, because he can't try more than one activity in a single day.
In the third testcase, Monocarp can choose day 2 for skiing, day 3 for a movie and day 7 for board games. 19+19+17=55 friends in total.
In the fourth testcase, Monocarp can choose day 1 for skiing, day 4 for a movie and day 9 for board games. 17+19+20=56 friends in total.
思路：
只要是不同的三天就可以，这样的话，我们只需要将a中人数由大到小的下标取前3，b中人数由大到小取前3，c中从大到小取前3，然后枚举，一共是 种情况其中再去掉一些重复的，为什么要取前3呢，假设说a这项活动把第一天占用了，对于b这项活动，第一天不能用了，还有第二大和第三大，对于c这项活动如果它的前两大的下标都和a,b重复了，他还可以取第三大，因此几乎总是可以取每个数组中的最大值。而当你不能取最大值时，就不需要考虑很多较小的数字。特别是，只需考虑每个数组中最大的三个数即可。
代码实现：
#include<bits/stdc++.h>
using namespace std;
void solve()
{
int n;
cin>>n;
vector<int> a(n),b(n),c(n);
for(int i=0;i<n;i++)
{
cin>>a[i];
}
for(int i=0;i<n;i++)
{
cin>>b[i];
}
for(int i=0;i<n;i++)
{
cin>>c[i];
}
vector<int> pa(n),pb(n),pc(n);
iota(pa.begin(),pa.end(),0);
iota(pb.begin(),pb.end(),0);
iota(pc.begin(),pc.end(),0);
sort(pa.begin(),pa.end(),
[&](int i,int j)
{
return a[i]>a[j];
});
sort(pb.begin(),pb.end(),
[&](int i,int j)
{
return b[i]>b[j];
});
sort(pc.begin(),pc.end(),
[&](int i,int j)
{
return c[i]>c[j];
});
int ans=0;
for(int i=0;i<3;i++)
{
for(int j=0;j<3;j++)
{
for(int k=0;k<3;k++)
{
int x=pa[i],y=pb[j],z=pc[k];
if(x!=y && x!=z && y!=z)
{
ans=max(ans,a[x]+b[y]+c[z]);
}
}
}
}
cout<<ans<<endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
