Billboard+hud+线段树单点更新

最新推荐文章于 2017-05-10 13:59:54 发布

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最新推荐文章于 2017-05-10 13:59:54 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/u012870383/article/details/38645173

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本文介绍了一个简单的单点更新算法，用于解决公告如何在限定尺寸的矩形公告板上布局的问题。该算法通过记录并更新每个区间的剩余宽度来确定公告的最佳放置位置，当无法放置时输出-1。

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Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10965 Accepted Submission(s): 4865

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

Sample Output

  
  
   
   1
2
1
3
-1
  
  
  
  
   
   解决方案：很简单的一道单点更新，只要在更新的时候随便搞搞。用一个数组记录每个区间的剩余宽度，贴广告时可根据每个区间的剩余宽度选择哪个位置贴，并更新该位置的剩余宽度，若找不到相应的位置输出-1。
  
  
  
  
   
   code:
   
   #include<iostream>
#include<cstdio>
#include<cstring>
#define MMAX 1200003
using namespace std;
int width[4*MMAX];
int n,h,w;
void build(int rt,int L,int R)
{

    if(L==R)
    {
        width[rt]=w;
    }
    else
    {
        int M=(L+R)/2;
        build(rt*2,L,M);
        build(rt*2+1,M+1,R);
        width[rt]=max(width[rt*2],width[rt*2+1]);
    }
}
void update(int rt,int L,int R,int W)
{

    int l=rt*2,r=rt*2+1;
    int M=(L+R)/2;
    if(L==R)
    {
        if(width[rt]<W)
        {
            printf("-1\n");
            return ;
        }
        width[rt]-=W;
        printf("%d\n",L);
        return ;
    }
    if(W<=width[l])
    {
        update(l,L,M,W);
    }
    else if(W<=width[r])
    {
        update(r,M+1,R,W);
    }
    else
    {
        printf("-1\n");
        return ;
    }
    width[rt]=max(width[l],width[r]);

}
int main()
{
    while(~scanf("%d%d%d",&h,&w,&n))
    {
        int N=min(h,n);
        build(1,1,N);
        for(int i=0; i<n; i++)
        {
            int a;
            scanf("%d",&a);
            update(1,1,N,a);
        }
    }
    return 0;
}