Billboard （线段树，多维）

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 524333

 

Sample Output

1213-1

题目大概：

有一个矩形w*h，沿着左上角放小矩形（高都为1），每放一个矩形，输出它在哪一行。

思路：

线段树的每个点是行数，每个点记录的是本行所剩宽度，范围内记录的是范围内的最大宽度。然后每次更新查询到最小的行数，并更新宽度，输出行数。

代码：

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;int sum[222222<<2];int h,w,n;void pushup(int rt){    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);}void build(int l,int r,int rt){   sum[rt]=w;    if(l==r)    {       return;    }    int m=(l+r)>>1;    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);}int quert(int x,int l,int r,int rt){    if(l==r)    {        sum[rt]-=x;        return l;    }    int m=(l+r)>>1;    int end;    if(sum[rt<<1]>=x){end=quert(x,l,m,rt<<1);}    else {end=quert(x,m+1,r,rt<<1|1);}    pushup(rt);    return end;}int main(){   while(~scanf("%d%d%d",&h,&w,&n))   {       if(h>=n)h=n;       build(1,h,1);       while(n--)       {           int x;           scanf("%d",&x);           if(sum[1]<x)printf("-1\n");           else printf("%d\n",quert(x,1,h,1));       }   }    return 0;}