P2085 min function value (minval), p2085minval

P2085 min function value (minval), p2085minval
Description

There are n functions, namely F1, F2,..., Fn. Define Fi (x) = AiX ^ 2 + BiX + Ci (x ε N *). Given these Ai, Bi, and Ci, the minimum m of all function values of all functions are requested (if there are duplicates, multiple functions need to be output ).

Input/Output Format Input Format:

Input data: enter two positive integers n and m in the first row. The following n rows have three positive integers in each row, where the three numbers in row I are Ai, Bi, and Ci respectively. Ai <= 10, Bi <= 100, Ci <= 10 000.

Output Format:

Output Data: output the first m elements after sorting all the function values generated by the n functions. The m number should be output to a row separated by spaces.

Input and Output sample Input example #1:

3 104 5 33 4 51 7 1

Output sample #1:

9 12 12 19 25 29 31 44 45 54

Description

Data scale: n, m <= 10000

 1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<queue> 5 using namespace std; 6 const int N = 1000000 + 5;  7 int a[N],b[N],c[N],p[N],n,m;  8 struct node 9 {10     int a,b;11 }f[N];12 int function(int i,int x)13 {14     return a[i]*x*x+b[i]*x+c[i];15 }16 void update(int i)17 {18     if(f[i].a>f[i*2].a && i*2<=n)19     {20         swap(f[i],f[i*2]);21         update(i*2);update(i/2);22     }23     if(f[i].a>f[i*2+1].a && i*2+1<=n)24     {25         swap(f[i],f[i*2+1]);26         update(i*2+1);update(i/2);27     }28 }29 int main()30 {31     scanf("%d%d",&n,&m);32     for(int i=1;i<=n;i++)33     {34         p[i]=1;35         scanf("%d%d%d",&a[i],&b[i],&c[i]);36         f[i].a=function(i,p[i]++);f[i].b=i;37     }38     for(int i=1;i<=n;i++)update(i);39     for(int i=0;i<m;i++)40     {41         printf("%d ",f[1].a);42         f[1].a=function(f[1].b,p[f[1].b]++);43         update(1);44     }45     return 0;46 }