HDU 2795 Billboard

Billboard

http://acm.hdu.edu.cn/showproblem.php?pid=2795

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5633    Accepted Submission(s): 2580

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

 

Author

hhanger@zju

 

Source

HDOJ 2009 Summer Exercise（5）

 

Recommend

lcy

 

 

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



const int N=200010;



#define L(rt) (rt<<1)

#define R(rt) (rt<<1|1)



struct Tree{

    int l,r;

    int len;    //该行长度(剩余长度,开始为w)

}tree[N<<2];



int h,w,n;



void PushUp(int rt){    //更新父结点的最大剩余长度

    tree[rt].len=max(tree[L(rt)].len,tree[R(rt)].len);

}



void build(int L,int R,int rt){

    tree[rt].l=L;

    tree[rt].r=R;

    tree[rt].len=w;

    if(tree[rt].l==tree[rt].r)

        return;

    int mid=(L+R)>>1;

    build(L,mid,L(rt));

    build(mid+1,R,R(rt));

    PushUp(rt);

}



int query(int val,int rt){

    if(tree[rt].l==tree[rt].r){     //找到适合的叶结点,并更新其剩余长度,输出其id号

        tree[rt].len-=val;

        return tree[rt].l;

    }

    int ans;

    if(val<=tree[L(rt)].len)    //当左右子树都同时满足的情况下，优先选择左子树，若发现len不满足则停止搜索这棵子树

        ans=query(val,L(rt));

    else if(val<=tree[R(rt)].len)

        ans=query(val,R(rt));

    PushUp(rt);

    return ans;

}



int main(){



    //freopen("input.txt","r",stdin);



    while(~scanf("%d%d%d",&h,&w,&n)){

        if(h>=n)    //n张海报最多张贴n行,多出无用

            h=n;

        build(1,h,1);

        int x;

        while(n--){

            scanf("%d",&x);

            if(tree[1].len>=x)  //1号结点是所有结点的父结点,若广告长度大于它的剩余长度,则无解

                printf("%d\n",query(x,1));

            else

                printf("-1\n");

        }

    }

    return 0;

}