D - 3N Numbers

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.

Input

Input is given from Standard Input in the following format:

Na1 a2 … a3N

Output

Print the maximum possible score of a'.

Sample Input 1

23 1 4 1 5 9

Sample Output 1

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

Sample Input 2

11 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.

Sample Input 3

38 2 2 7 4 6 5 3 8

Sample Output 3

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

1 #include 
        
          2 using namespace std; 3 #define LL long long 4 #define INF (1LL<<62) 5 #define MX 100005*3 6  7 LL a[MX]; 8 LL ma[MX]; 9 LL mi[MX];10 11 int main()12 {13     LL n;14     cin>>n;15 16     for (int i=1;i<=3*n;i++)17         scanf("%lld",&a[i]);18     priority_queue 
         
           Q;19     LL sum = 0;20     for (int i=1;i<=2*n;i++)21     {22         Q.push(-a[i]);23         sum+=a[i];24 25         if (i>n) sum += Q.top(),Q.pop();26         ma[i]=sum;27     }28 29     while (!Q.empty()) Q.pop();30     sum=0;31     for (int i=3*n;i>=n+1;i--)32     {33         Q.push(a[i]);34         sum+=a[i];35 36         if (i<=2*n) sum -= Q.top(),Q.pop();37         mi[i]=sum;38     }39 40     LL ans = -INF;41     for (int i=n;i<=2*n;i++)42         ans = max (ans, ma[i]-mi[i+1]);43     cout<
          
           <

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转载于:https://www.cnblogs.com/haoabcd2010/p/6884609.html

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