Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7320 Accepted Submission(s): 3281

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

【题解】：

构造线段树

struct Nod
{
int l,r; //叶子节点的l表示第几行h
int w; //剩余宽度
}node[N<<2];

【code】：

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <algorithm>
 5 #define N 200010
 6 using namespace std;
 7 
 8 struct Nod
 9 {
10     int l,r;   //叶子节点的l表示第几行h
11     int w;  //剩余宽度
12 }node[N<<2];
13 
14 int h, w, row;
15 
16 void building(int l,int r,int p)
17 {
18     node[p].l = l;
19     node[p].r = r;
20     node[p].w = w;
21     if(l+1==r)  return;  //叶子
22     int mid = (l+r)>>1;
23     building(l,mid,p<<1);
24     building(mid,r,p<<1|1);
25 }
26 
27 void update(int p,int w0)
28 {
29     if(node[p].w<w0)    return;
30     if(node[p].l+1==node[p].r)  //找到叶子节点
31     {
32         row = node[p].l;
33         node[p].w-=w0;
34         return;
35     }
36     if(node[p<<1].w>=w0)    update(p<<1,w0);  //往左查找
37     else if(node[p<<1|1].w>=w0) update(p<<1|1,w0);  //往右
38     node[p].w = max(node[p<<1].w,node[p<<1|1].w);  //向上更新
39 }
40 
41 int main()
42 {
43     int n;
44     while(~scanf("%d%d%d",&h,&w,&n))
45     {
46         int m = min(h,n)+1;  //注意加1，因为第一个叶子为1,2  最后一个为 min(h,n)，min(h,n)+1
47         building(1,m,1);
48         int w0;
49         while(n--)
50         {
51             scanf("%d",&w0);
52             row = -1;
53             update(1,w0);
54             printf("%d
",row);
55         }
56     }
57 }