P1631 and p1631

P1631 and p1631
Description

There are two sequences A and B whose lengths are all N. In A and B, we can get N ^ 2 and N ^ 2 and the N ^ 2 and the N smallest N.

Input/Output Format

Input Format:

 

The first line is a positive integer N;

The second line contains N integers: Ai <= Ai + 1 and Ai <= 10 ^ 9;

The third line contains N integer Bi, meeting the requirements of Bi <= Bi + 1 and Bi <= 10 ^ 9.

[Data scale]

1 <= N <= 50%;

For 100% of the data, 1 <=n <= 100000.

 

Output Format:

 

The output contains only one row and N integers. The smallest sum of N values is output from small to large. Adjacent numbers are separated by spaces.

 

Input and Output sample input sample #1:

32 6 61 4 8

Output sample #1:

3 6 7

Note that
Priority_queue is slower than heap
Make_heap is slower than pop_heap

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 const int MAXN=1000001; 9 priority_queue<int>q;10 void read(int & n)11 {12     char c='+';int x=0;int flag=0;13     while(c<'0'||c>'9')14     {15         c=getchar();16         if(c=='-')flag=1;17     }18     19     while(c>='0'&&c<='9')20         x=x*10+c-48,c=getchar();21     if(flag==1)n=-x;22     else n=x;23 }24 int n,tot;25 int a[MAXN],b[MAXN],c[MAXN];26 int main()27 {28     read(n);29     for(int i=1;i<=n;i++)30     read(a[i]);31     for(int i=1;i<=n;i++)32     read(b[i]);33     for(int i=1;i<=n;i++)34         c[i]=a[1]+b[i];35     make_heap(c+1,c+n+1);36     for(int i=2;i<=n;i++)37     {38         for(int j=1;j<=n;j++)39         {40             if(a[i]+b[j]<c[1])41             {42                 //q.push(a[i]+b[j]);43                 c[1]=a[i]+b[j];44                 pop_heap(c+1,c+n+1);45             }46             else break;47         }48     }49     //priority_queue<int,vector<int>,greater<int> >p;50     make_heap(c+1,c+n+1,greater<int>());51     int k=n;52     for(int i=1;i<=n;i++)53     {54         printf("%d ",c[1]);55         pop_heap(c+1,c+1+k,greater<int>());56         k--;57     }58     return 0;59 }