HDU 2795 Billboard 线段树维护区间最大值&&查询变形

Posted 2021-01-02 ymzjj

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

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28743    Accepted Submission(s): 11651

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

[email protected]

 

 

Source

HDOJ 2009 Summer Exercise（5）

 

题意概括：

有一个 高为H 宽为W 的公告墙，要在上面贴 N 个 高恒为1 宽为 W 的广告（每次都是优先左上方向)，输出广告贴的高度 h； 

解题思路：

线段树的子节点维护 1~H 每一层的所剩的最大值 Wi，每次查询不是找最大值，而是找到最大值所在的结点位置（第几层）

Tip：模板是死的，数据结构是活的。

Push操作从叶子结点往上更新；

线段树最大的边界不一定是输入的 H ，也可能是 N（极限每一层只贴一个）

但用min会超时很多，淳朴的 if 语句判断 H 还是 N 小就好

查询先遍历左子树后遍历右子树，因为优先贴左边。

看似简单的东西要自己亲手去敲代码才能发现问题所在。

AC code：

 1 #include <bits/stdc++.h>
 2 #define lson l, mid, root<<1
 3 #define rson mid+1, r, root<<1|1
 4 using namespace std;
 5 
 6 const int MAXN = 2e5+10;
 7 int maxh[MAXN<<2], hb;
 8 int H, W, N;
 9 int read()
10 {
11     int f=1,x=0;char s=getchar();
12     while(s<‘0‘||s>‘9‘){if(s==‘-‘)f=-1;s=getchar();}
13     while(s>=‘0‘&&s<=‘9‘){x=x*10+s-‘0‘;s=getchar();}
14     x*=f;
15     return x;
16 }
17 void Build(int l, int r, int root)
18 {
19     maxh[root] = W;
20     if(l == r) return;
21     //int mid = l+((r-l)>>1);
22     int mid = (l+r)>>1;
23     Build(lson);Build(rson);
24 }
25 void Push(int Root)
26 {
27     maxh[Root] = max(maxh[Root<<1], maxh[Root<<1|1]);
28 }
29 
30 void Update(int pos, int l, int r, int root, int len)
31 {
32     if(l == r){maxh[root]+=len;return;}
33     //int mid = l+((r-l)>>1);
34     int mid = (l+r)>>1;
35     if(pos <= mid) Update(pos, lson, len);
36     else Update(pos, rson, len);
37     Push(root);
38 }
39 int Query(int l, int r, int root)
40 {
41     if(l == r) return l;
42     int mid = (l+r)>>1;
43     int res = 0;
44     if(maxh[root<<1] >= hb) res = Query(lson);
45     else res = Query(rson);
46     return res;
47 }
48 int main()
49 {
50     while(~scanf("%d %d %d", &H, &W, &N)){
51         if(H > N) H = N;
52         Build(1, H, 1);
53         for(int i = 1; i <= N; i++)
54         {
55             hb = read();
56             //scanf("%d", &hb);
57             if(maxh[1] < hb) printf("-1
");
58             else{
59                 int now = Query(1, H, 1);
60                 Update(now, 1, H, 1, -hb);
61                 printf("%d
", now);
62             }
63         }
64     }
65     return 0;
66 }

View Code