AtCoder Regular Contest 074 D.3N Numbers

D - 3N Numbers

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

Let N be a positive integer.

There is a numerical sequence of length 3N, a=(a1,a2,…,a3N). Snuke is constructing a new sequence of length 2N, a', by removing exactly N elements from a without changing the order of the remaining elements. Here, the score of a' is defined as follows: (the sum of the elements in the first half of a')−(the sum of the elements in the second half of a').

Find the maximum possible score of a'.

Constraints

• 1≤N≤105
• ai is an integer.
• 1≤ai≤109

Partial Score

• In the test set worth 300 points, N≤1000.

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Input

Input is given from Standard Input in the following format:

N
a1 a2 … a3N

Output

Print the maximum possible score of a'.

 把3N数组分为两部分，前部分选择N个最大的元素，后部分选择N个最小的元素，利用优先队列求一下和即可

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define max(x,y) (x)>(y)?(x):(y)
#define min(x,y) (x)>(y)?(y):(x)
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1<<28;
const int maxn=100010;
const int maxm=110;
const int maxk=100010;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n;
int num[maxn*3];
ll maxleft[maxn],minright[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    int a,b;
    while(~scanf("%d",&n))
    {
        cls(maxleft,0);
        cls(minright,0);
        for(int i=0;i<n*3;i++)
            scanf("%d",&num[i]);
        priority_queue<int, vector<int>, greater<int> > Q1;
        priority_queue<int> Q2;
        for(int i=0;i<n;i++)
        {
            maxleft[0]+=num[i];
            Q1.push(num[i]);
            minright[n+1]+=num[n*3-1-i];
            Q2.push(num[n*3-1-i]);
        }
        for(int i=1;i<=n;i++)
        {
            Q1.push(num[n+i-1]);
            maxleft[i]=maxleft[i-1]+num[n+i-1]-Q1.top();
            Q1.pop();
 
            Q2.push(num[n*2-i]);
            minright[n+1-i]=minright[n+1-i+1]+num[n*2-i]-Q2.top();
            Q2.pop();
        }
        ll ans;
        for(int i=0;i<=n;i++)
        {
            if(i==0)
                ans=maxleft[i]-minright[i+1];
            else
                ans=max(ans,maxleft[i]-minright[i+1]);
        }
        printf("%lld
",ans);
    }
    return 0;
}