HDU2795Billboard--线段树单点更新

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7538    Accepted Submission(s): 3348

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

 

题意:h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子

思路:每次找到最大值的位子,然后减去L；

感觉hh线段树的写法灰常漂亮，这是我自己的写的。。有点像，哈哈。。。

代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;
 

const int maxn=200000;
 

struct Tree
{
    int l;
    int r;
    int sum;//表示剩余空间的最大值；
}T[maxn<<2];
 

int h,w,n,ans,tmp;
 

void pushup(int u)
{
    T[u].sum=max(T[u<<1].sum,T[u<<1|1].sum);
}
 

void build(int l,int r,int u)
{
    T[u].l=l,T[u].r=r;
    if(l==r)
    {
        T[u].sum=w;
        return;
    }
    int mid=(T[u].l+T[u].r)>>1;

    build(l,mid,u<<1);
    build(mid+1,r,u<<1|1);

    pushup(u);
}
 

void update(int pos,int value,int u)
{
    if(T[u].l==T[u].r)
    {
        T[u].sum-=value;
        return;
    }

    int mid=(T[u].l+T[u].r)>>1;
    if(pos<=mid)
        update(pos,value,u<<1);
    else
        update(pos,value,u<<1|1);

    pushup(u);//每次更新不仅更新父节点，还要更新子节点；
}
 

int query(int l,int r,int u,int value)
{
    if(T[u].sum<value)
    return -1;

    if(T[u].l==T[u].r)
    return T[u].l;

    int mid=(T[u].l+T[u].r)>>1;
    if(T[u<<1].sum>=value)
        return query(l,r,u<<1,value);
    else
        return query(l,r,u<<1|1,value);
}
 

int main()
{
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        int m=min(h,n);
        build(1,m,1);

        while(n--)
        {
            scanf("%d",&tmp);
            ans=query(1,m,1,tmp);
            printf("%d\n",ans);
            if(ans!=-1)
                update(ans,tmp,1);
        }
    }
    return 0;
}
 

这块一般注意数组的范围不然容易RE，幸亏我开了20W，一遍AC，哈哈。