Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17190 Accepted Submission(s): 7266
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 would always
choose the topmost possible position for the announcement. Among all
possible topmost positions she would 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 participants
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.
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.
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.
2
4
3
3
3
2
1
3
-1
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=;
int t[maxn<<];
int h,w,n,x; void Build(int node,int l,int r)
{
t[node]=w;
if(l==r)return;
int mid=(l+r)>>;
Build(node<<,l,mid);
Build(node<<|,mid+,r);
} void Query(int node,int l,int r,int q)
{
if(l==r){
printf("%d\n",l);
t[node]-=q;
return;
}
int mid=(l+r)>>,a=node<<,b=a+;
if(t[a]>=q)Query(a,l,mid,q);
else Query(b,mid+,r,q);
t[node]=max(t[a],t[b]);
return;
}
int main()
{
while(~scanf("%d%d%d",&h,&w,&n)){
h=min(n,h);
Build(,,h);
for(int i=;i<=n;i++){
scanf("%d",&x);
if(x>t[])
printf("-1\n");
else
Query(,,h,x);
}
}
return ;
}