Billboard

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20145    Accepted Submission(s): 8391

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）

 

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/*
题意：有一块h*w的空广告牌，现在n个1*w的广告要贴上去，现在问

初步思路：实际上就是，一块一块的广告向上贴，但是贴之前查询一下，前面的已经贴过的位置是不是能盛下这个广告牌，如果能盛下的话
    就放上去，如果不能放下的话，就重新向下找位置贴，实际的操作：从1开始到n找第一个大于wi的位置，然后返回位置，然后将wi更新到
    到这个位置，这样的话每个几点就得保存区间最大值了，不能保存和了
*/
#include<bits/stdc++.h>
using namespace std;
int w,h,n;
int a;
int now=0;//表示当前开扩的广告范围
/****************************线段树基础模板*********************************/
const int maxn=200000+10;

int sum[maxn*4];
#define lson i*2, l, m
#define rson i*2+1, m+1, r

void PushUp(int i)
{
    sum[i]=max(sum[i*2],sum[i*2+1]);
}

void build(int i,int l,int r)
{
    if(l==r)
    {
        sum[i]=w;//初始化这个位置没有广告，所以空位置为w
        return ;
    }
    int m=(l+r)/2;
    build(lson);
    build(rson);
    PushUp(i);
}
//查询改为，查询从1开始的第一个比val大的数返回坐标
int query(int val,int i,int l,int r)
{
    // cout<<l<<" "<<r<<endl;
    if(l==r)//如果遍历到根节点的话，就可以输出了
        return l;
    int m=(l+r)/2;
    //这句话才能体现从1开始找,肯定是优先遍历坐左儿子
    if(sum[i*2]>=val) query(val,lson);
    else if(sum[i*2+1]>=val) query(val,rson);
}

void update(int id,int val,int i,int l,int r)
{
    if(l==r)
    {
        sum[i]+=val;
        return ;
    }

    int m=(l+r)/2;
    if(id<=m) update(id,val,lson);
    else update(id,val,rson);
    PushUp(i);
}

/****************************线段树基础模板*********************************/
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&h,&w,&n)!=EOF){
        now=min(h,n);
        build(1,1,now);//初始化建树
        for(int i=0;i<n;i++){
            scanf("%d",&a);
            // cout<<"sum[1]="<<sum[1]<<endl;
            if(sum[1]<a){
                puts("-1");
                continue;
            }
            int res=query(a,1,1,now);
            printf("%d
",res);
            update(res,-a,1,1,now);
        }
    }
    return 0;
}