单点修改，线段树维护区间最大值——广告牌

最新推荐文章于 2021-02-26 23:30:41 发布

原创最新推荐文章于 2021-02-26 23:30:41 发布 · 271 阅读

1 ·

CC 4.0 BY-SA版权

线段树专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种用于在大型矩形公告牌上高效布局公告的算法。该算法通过使用线段树数据结构，使得每次都能找到最高且最左侧的可用位置来放置新的公告，即使在高度限制极大的情况下也能有效处理多达20万条公告。文章提供了详细的实现代码，包括如何预处理以适应大量公告的快速查找和更新。

Billboard

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

Sample Output

题意每块广告优先选择最高的地方粘贴；

做法把弄一个1-h的线段树，广告宽度就是每个区间的最大值；

对于每一个新加入的广告，优先向左找，每个结点是区间的最大值；

因为h太大了，但是最多n个广告，所以预处理一下就好了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
const int maxn=3e7+7;
int h,w,n;
int b[maxn];
int ans;
void build(int l,int r,int rt)
{
    if(l==r)
    {
        b[rt]=w;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt*2);
    build(mid+1,r,rt*2+1);
    b[rt]=max(b[rt*2],b[rt*2+1]);
}
void change(int l,int r,int rt,int v)
{
    if(l==r)
    {
        b[rt]-=v;
        cout<<l<<endl;
        return ;
    }
    int mid=(l+r)>>1;
    if(b[rt*2]>=v)    change(l,mid,rt*2,v);
    else    change(mid+1,r,rt*2+1,v);
    b[rt]=max(b[rt*2],b[rt*2+1]);

}
int main()
{

    int v;
    while(~scanf("%d%d%d",&h,&w,&n))
{
    if(h>n) h=n;
    build(1,h,1);
    while(n--)
    {
        scanf("%d",&v);
        if(b[1]>=v) change(1,h,1,v);
        else cout<<-1<<endl;
    }
    }
    return 0;
}