题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2795
Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13050 Accepted Submission(s):
5651
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.
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.
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.
#include <iostream>
#include <cstdio> using namespace std; struct node
{
int l,r;
int mmax;
} s[*]; void InitTree(int l,int r,int k,int w)
{
s[k].l=l;
s[k].r=r;
s[k].mmax=w;
if (l==r)
return ;
int mid=(l+r)/;
InitTree(l,mid,*k,w);
InitTree(mid+,r,*k+,w);
} void UpdataTree(int num,int k)//进行点的更新
{
if (s[k].r==s[k].l)
{
printf ("%d\n",s[k].l);
s[k].mmax-=num;
return ;
}
if (num<=s[k*].mmax)
UpdataTree(num,k*);
else
UpdataTree(num,k*+);
s[k].mmax=s[k*].mmax>s[k*+].mmax?s[k*].mmax:s[k*+].mmax;
} int main ()
{
int h,w,n,a;
while (~scanf("%d%d%d",&h,&w,&n))
{
int hh=h>n?n:h;
InitTree(,hh,,w);
for (int i=; i<=n; i++)
{
scanf("%d",&a);
if (a<=s[].mmax)
UpdataTree(a,);
else
printf ("-1\n");
}
}
return ;
}