【hdu2795】Billboard

Problem Description

At 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.

 

Input

There 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.

 

Output

For 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.

 

Sample Input

3 5 5
2
4
3
3
3

 

Sample Output

1
2
1
3
-1

 

Author

hhanger@zju

 

Source

HDOJ
2009 Summer Exercise（5）

题目大意：给定h*w的广告牌,让你每次把给定长度1*wi的广告尽量往高再尽量往左贴，不能分割，输出位置（h），不能贴就输出-1。

思路：储存最大值，若左儿子大于长度就到左节点，否则就到右节点，确定为哪一行时减去输出即可。

attention:注意数据范围,以w为节点，h为节点数建树，但因为”1 <= h,w <= 10^9“h一大必然boom shakalaka，所以h>n时取n即可，因为即使每个wi都等于w也可以满足，wi比w大，h再多也没用，因为不能劈成两半，且1<=n<=200000。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #define MAXN 222222
 using namespace std;
 int segtree[MAXN*];
 int n,h,t,x,ans;
 void adddata(int now)
 {
     segtree[now]=max(segtree[(now<<)],segtree[(now<<)+]);
 }
 void buildtree(int now,int l,int r)
 {
     if (l==r)    {segtree[now]=t;    return;}
     int mid=(l+r)>>;
     buildtree((now<<),l,mid);
     buildtree((now<<)+,mid+,r);
     adddata(now);
 }
 int query(int now,int l,int r,int v)
 {
     if (l==r)    {segtree[now]-=v;return l;}
     int mid=(l+r)>>;
     if (segtree[(now<<)]>=v)    ans=query((now<<),l,mid,v);
     else    ans=query((now<<)+,mid+,r,v);
     adddata(now);
     return ans;
 }
 int main()
 {
     int i,p;
     while (~scanf("%d%d%d",&h,&t,&n))
     {
         h=min(h,n);
         buildtree(,,h);
         for (i=;i<=n;i++)
         {
             scanf("%d",&x);
             if (x>segtree[])    printf("-1\n");
             else{    p=query(,,h,x);    printf("%d\n",p);}
         }
     }
     return ;
 }