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 3Sample Output
1 2 1 3 -1题意： 一个h*w的木板，n次小木板放置，大小1*wi,尽量把小木板放在大木板的左上方，如果能放输出放在第几行，如果不能就输出-1
思路：因为小木板h为1，所以就用大木板的h建线段树，为了避免h特别大而n很小，根本用不完h这种情况，所以h=min(h,n)；
用数组idx来记录当前线段树的根结点为第几排 ；因为要求尽可能往上放，所以先判断左子树，再判断右子树
import java.io.BufferedReader; import java.io.InputStreamReader; import java.math.BigDecimal; import java.math.BigInteger; import java.util.Arrays; import java.util.StringTokenizer; class InputReader2 { BufferedReader buf; StringTokenizer tok; InputReader2() { buf = new BufferedReader(new InputStreamReader(System.in)); } boolean hasNext() { while(tok == null || !tok.hasMoreElements()) { try { tok = new StringTokenizer(buf.readLine()); } catch(Exception e) { return false; } } return true; } String next() { if(hasNext()) return tok.nextToken(); return null; } int nextInt() { return Integer.parseInt(next()); } } public class Main { static int h,w,n; static final int max=200005; static int tree[]=new int[max<<2]; static int idx[]=new int[max<<2]; static int cnt; public static void build(int node,int l,int r){ if(l==r){ tree[node]=w; idx[node]=++cnt;// 用数组idx来记录当前线段树的根结点为第几排 return; } int mid=(l+r)>>1; build(node<<1,l,mid); build(node<<1|1,mid+1,r); tree[node]=Math.max(tree[node<<1], tree[node<<1|1]); } public static int update(int node,int l,int r,int a){ if(l==r){ tree[node]-=a; return idx[node]; } int ans=-1; int mid=(l+r)>>1; //因为要求尽可能往上放，所以先判断左子树，再判断右子树 //一定是if else if ，只要有一个满足就行 if(tree[node<<1]>=a) ans=update(node<<1,l,mid,a); else if(tree[node<<1|1]>=a) ans=update(node<<1|1,mid+1,r,a); tree[node]=Math.max(tree[node<<1], tree[node<<1|1]); return ans; } public static void main(String[] args) { InputReader2 scan=new InputReader2(); while(scan.hasNext()){ cnt=0; Arrays.fill(tree, 0); Arrays.fill(idx, 0); h=scan.nextInt(); w=scan.nextInt(); n=scan.nextInt(); int min=Math.min(h, n);//为了避免h特别大而n很小，根本用不完h这种情况，所以h=min(h,n) build(1,1,min); while(n!=0){ int a=scan.nextInt(); if(tree[1]<a){ System.out.println("-1"); } else{ System.out.println(update(1,1,min,a)); } n--; } } } }
