Summary of the number of line segments (1) HDU 2795 Billboard !!

Billboard time limit: 20000/8000 ms (Java/other) memory limit: 32768/32768 K (Java/other) total submission (s): 6 accepted submission (s): 3 Font :{
Profont ('times new Roman ')
} "> Times New Roman | {
Profont ('verdana ')
} "> Verdana | {
Profont ('Georgia ')
} "> Georgia font size :{
Profontadd (-1)
} "> Detail {
Profontadd (1)
} "> → {
Objfolder ('procon ')
} "> Problem descriptionat the entrance to the university, there is a huge rectangular billboard of size H * w (H is its height and W is its width ). the Board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the I-th announcement is a rectangle of size 1 * WI.

When someone puts a new announcement on the billboard, she wowould always choose the topmost possible position for the announcement. Among all possible topmost positions she wowould always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the Billboard (that's why some programming contests have no participant ants from this university ).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

{
Objfolder ('proinput ')
} "> Inputthere are multiple cases (no more than 40 cases ).

The first line of the input file contains three integer numbers, H, W, and N (1 <= H, W <= 10 ^ 9; 1 <= n <= 200,000) -The dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10 ^ 9)-the width of I-th announcement.

{
Objfolder ('prooutput ')
} "> Outputfor each announcement (in the order they are given in the input file) output one number-the number of the row in which this announcement is placed. rows are numbered from 1 to H, starting with the top row. if an announcement can't be put on the billboard, output "-1" for this announcement. {
Objfolder ('prosamplein ')
} "> Sample input

3 5 524333

{
Objfolder ('prosampleout ')
} "> Sample output

1213-1

The question time limit is 8 seconds, but there is an ominous hunch! The meaning of the question is easy to understand, but the true understanding is clever !! The question refers to posting a newspaper on the bulletin board,

On the left side of the same height, the newspaper is very standard. The unit length is high !!!!!! You can see this!

H, W, and N (1 <= H, W <= 10 ^ 9; 1 <= n <= 200,000)

H is so big that the line segment tree cannot be built !! But look at N, not big !! That's useless if H is so big! Think about it! H is bigger than n. I Can't paste a row of posters! Therefore

You only need to create a seg_tree [3*200000] with enough length !!

Not much! Post Code for your understanding!

#include<iostream>
using namespace std;

const __int64 N=200000;
struct seg_tree
{
 int left,right;
 int size;
}t[4*N];
inline int MID(int a,int b)
{
 return (a+b)/2;
}
inline int MAX(int a,int b)
{
 return a>b?a:b;
}
int n,h,w,flag;

void make_tree(int c,int l,int r)
{
 t[c].left = l;
 t[c].right = r;
 t[c].size = w;
 if(l == r)
 {
  return;
 }
 int mid = MID(l,r);
 make_tree (2*c,l,mid );
 make_tree(2*c+1,mid+1,r);
}

void find_ans(int c,int val)
{
 if(val > t[c].size )
 {
  return;
 }
 if(t[c].left==t[c].right)
 {
  t[c].size -= val;
  flag = t[c].left;
  return;
 }
 if(t[2*c].size >= val)
  find_ans(c*2,val);
 else
  find_ans(2*c+1,val);
 t[c].size = MAX(t[c*2].size,t[c*2+1].size);
}

int main()
{
 int i,temp;
 while(cin>>h>>w>>n)
 {
  h = h<n?h:n;
  make_tree(1,1,h);
  for(i=1;i<=n;i++)
  {
   flag = 0;
   scanf("%d",&temp);
   find_ans(1,temp);
   if(flag!=0)
    printf("%d/n",flag);
   else
    printf("-1/n");
  }
 }
 return 0;
}