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'.

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Sample Input 1

2
3 1 4 1 5 9

Sample Output 1

1

When a2 and a6 are removed, a' will be (3,4,1,5), which has a score of (3+4)−(1+5)=1.

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Sample Input 2

1
1 2 3

Sample Output 2

-1

For example, when a1 are removed, a' will be (2,3), which has a score of 2−3=−1.

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Sample Input 3

3
8 2 2 7 4 6 5 3 8

Sample Output 3

5

For example, when a2, a3 and a9 are removed, a' will be (8,7,4,6,5,3), which has a score of (8+7+4)−(6+5+3)=5.

/// 题意是:有一个 3n 长的序列，现拿走 n 个数，然后分成前 n 个数，和后 n 个数 ，求前n个数和减后 n 个数和的最大值

// 用一个优先队列保存区间最大 n 数和，并赋给数组保存

用一个优先队列保存区间最小 n 数和，并赋给数组保存

最后循环一遍即可

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define INF (1LL<<62)
#define MX 100005*3

LL a[MX];
LL ma[MX];
LL mi[MX];

int main()
{
    LL n;
    cin>>n;

    for (int i=1;i<=3*n;i++)
        scanf("%lld",&a[i]);
    priority_queue <LL> Q;
    LL sum = 0;
    for (int i=1;i<=2*n;i++)
    {
        Q.push(-a[i]);
        sum+=a[i];

        if (i>n) sum += Q.top(),Q.pop();
        ma[i]=sum;
    }

    while (!Q.empty()) Q.pop();
    sum=0;
    for (int i=3*n;i>=n+1;i--)
    {
        Q.push(a[i]);
        sum+=a[i];

        if (i<=2*n) sum -= Q.top(),Q.pop();
        mi[i]=sum;
    }

    LL ans = -INF;
    for (int i=n;i<=2*n;i++)
        ans = max (ans, ma[i]-mi[i+1]);
    cout<<ans<<endl;
    return 0;
}