Hdu 2795 Billboard, hdu2795billboard

Hdu 2795 Billboard, hdu2795billboard
BillboardTime Limit: 20000/8000 ms (Java/Other) Memory Limit: 32768/32768 K (Java/Other) Total Submission (s): 47 Accepted Submission (s ): 18 Problem DescriptionAt 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 wowould always choose the topmost possible position for the announcement. Among all possible topmost positions she wowould 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 participant ants 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.

 

InputThere 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.

 

OutputFor 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 Input3 5 524333

 

Sample Output1213-1

 

Authorhhanger @ zju

 

SourceHDOJ 2009 Summer Exercise (5): First of all, it is difficult to build a project. I think of several solutions, because they are all attached with a height of 1. 1. Building by area is obviously incorrect. 2. Build by height. A problem that needs to be paid attention to is that the maximum height is 1e9. Therefore, we must judge the size of him and n. At first, we didn't pay attention to this, so we kept RE and thought it was a small array. You can also directly determine the maximum value of the root node. If it is put into updata, there are many more errors. A lot of problems are hard to think about. After thinking about it, I found that the problem was just separated by a window.

# Include <iostream> # include <cstring> # include <cmath> # include <cstdio> # include <queue> using namespace std; int h, w, o; struct node {int l, r, v; // l table height, maximum value that can be added to table v} tr [200100 <2]; void build (int n, int l, int r) {tr [n]. l = l; tr [n]. r = r; tr [n]. v = w; if (tr [n]. l = tr [n]. r) return; int mid = (l + r)> 1; build (n <1, l, mid); build (n <1 | 1, mid + 1, r);} int updata (int v, int n) {int res; if (tr [n]. l = tr [n]. r) {tr [n]. v-= v; return tr [n ]. L;} if (tr [n <1]. v> = v) res = updata (v, n <1); else res = updata (v, n <1 | 1); tr [n]. v = max (tr [n <1]. v, tr [n <1 | 1]. v); return res;} int main () {while (~ Scanf ("% d", & h, & w, & o) {build (1, 1, min (h, o )); for (int I = 0; I <o; I ++) {int x; scanf ("% d", & x); if (tr [1]. v <x) printf ("-1 \ n"); else printf ("% d \ n", updata (x, 1 ));}}}