hdu 2795 Billboard（线段树-剩余空间）

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 524333 

题意就是有一块模板高h宽w，然后有很多海报要往上面贴，每张海报宽wi高1，要贴时都会自顶向下搜索，一旦发现有一行长度足够就会贴到这一行最左边还没被贴到的位置

输出贴上去的海报所在的行数，没地方贴输出-1

用线段树维护该区间剩余的最大长度，底部节点则为每一行的剩余长度，用区间搜索寻找合适的位置即可。

本题的一个坑点在于线段树的区间最大值，本来以为要开到10^9,可是这样必然会爆掉，其实只需要开到海报张数即可，当然如果高度比海报张数少则开到高度即可。

#include<stdio.h>#define lson l,m,2*rt#define rson m+1,r,2*rt+1const int MAXN=200001;int lave[MAXN*4];int w;void pushup(int rt){lave[rt]=lave[rt*2]>lave[rt*2+1]?lave[rt*2]:lave[rt*2+1];}void build(int l,int r,int rt){if(l==r){lave[rt]=w;return ;}int m=(l+r)/2;build(lson);build(rson);pushup(rt);}void paste(int len,int l,int r,int rt){if(l==r){lave[rt]-=len;printf("%d\n",l);return ;}int m=(l+r)/2;if(len<=lave[rt*2]) paste(len,lson);else paste(len,rson);pushup(rt);}int main(){int h,n,len;while(scanf("%d%d%d",&h,&w,&n)!=EOF){h=h>n?n:h;build(1,h,1);while(n--){scanf("%d",&len);if(len>lave[1]) printf("-1\n");else paste(len,1,h,1);}}return 0;}