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

題意是有一塊長h寬w的板子，你可以往上貼廣告，位置優先越靠上。
max維護一個剩餘最大的空間，若最大空間Max[1]小於廣告的寬度，則直接返回-1.
線段樹搜尋第一個大於等於w值的位置，query和update寫一起。
注意h的值，線段樹開的空間應該根據n來開。
而且輸入規模過大，不要用Scanner，用BufferedReader輸入。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main{
    public
 
 static int[] max;
    public static void pushUp(int rt){
        max[rt]=Math.max(max[rt<<1],max[rt<<1|1]);
    }
    public static void build(int l,int r,int rt,int val){
        max[rt]=val;
        if(l==r)return;
        int mid=(l+r)>>1;
        build(l,mid,rt<<1,val);
        build(mid+1
 
,r,rt<<1|1,val);
    }
    public static int query(int w,int l,int r,int rt){
        if(l==r){
            max[rt]-=w;
            return l;
        }
        int mid=(l+r)>>1;
        int ret;
        if(max[rt<<1]>=w) {
            ret=query(w,l,mid,rt<<1);
        }else{
            ret=query(w,mid+1,r,rt<<1|1);
        }
        pushUp(rt);
        return ret;
    }
    public static void main(String[] args) throws IOException {
        //Scanner cin=new Scanner(System.in);
        BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
        while (true){
            String[] tmp;
            try{
                tmp=reader.readLine().split("\\s+");
            }catch (Exception e){
                return;
            }
            int h,w,n;
            h=Integer.parseInt(tmp[0]);
            w=Integer.parseInt(tmp[1]);
            n=Integer.parseInt(tmp[2]);
            h=Math.min(h,n);
            if(h!=1){
                max=new int[h<<2];
            }else{
                max=new int[10];
            }
            build(1,h,1,w);
            for(int i=1;i<=n;++i){
                int q=Integer.parseInt(reader.readLine());
                if(max[1]<q){
                    System.out.println("-1");
                }else {
                    System.out.println(query(q,1,h,1));
                }
            }
        }
    }
}