HDU 2795 Billboard (simple line segment tree)

HDU 2795 Billboard (simple line segment tree)
BillboardTime Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 12812 Accepted Submission (s): 5578


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 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.
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 524333

Sample Output

1213-1

Author hhanger @ zju
Source HDOJ 2009 Summer Exercise (5)
Recommend lcy | We have carefully selected several similar problems for you: 1698 1542 1828 1540


A: There is an h * w wooden board. To advertise on the board, you must stick it as much as possible. If not, stick it to the left as much as possible, (In this way, the number of ads can be the most.) each advertisement is 1 * wi, which means that each advertisement can only occupy one line. You can use the line segment tree for maintenance and compare each time, if the remaining width at the top is greater than or equal to the width of the advertisement, then the width of the advertisement on the dashboard is subtracted. If you can think of this method, it is very simple. If you can't think of it, you don't know how to do it...

#include
 
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       using namespace std;const int maxn = 200001;struct node{ int l; int r; int cnt;}q[maxn<<4];int h,w,n;void build(int l,int r,int rt){ q[rt].l = l; q[rt].r = r; q[rt].cnt = 0; if(q[rt].l == q[rt].r) { q[rt].cnt = w; return ; } int mid = (l+r)>>1; build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt);}int qurry(int k,int l,int r,int rt){ if(q[rt].l == q[rt].r) { q[rt].cnt -= k; return q[rt].l; } int mid = (l+r)>>1; int ans = 0; if(q[rt<<1].cnt>=k) { ans = qurry(k,l,mid,rt<<1); } else { ans = qurry(k,mid+1,r,rt<<1|1); } q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt); return ans;}int main(){ while(scanf("%d%d%d",&h,&w,&n)!=EOF) { if(h>n) { h = n; } build(1,h,1); while(n--) { int m; scanf("%d",&m); if(m>q[1].cnt) { printf("-1\n"); } else { int ans = qurry(m,1,h,1); printf("%d\n",ans); } } } return 0;}