Hdu2795 Billboard (line segment tree)

Hdu2795 Billboard (line segment tree)
BillboardTime Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 16647 Accepted Submission (s): 7037

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, changesin 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 Input

3 5 524333

Sample Output

1213-1

Authorhhanger @ zju
SourceHDOJ 2009 Summer Exercise (5)
Question: An advertisement is placed on an h * w advertisement board. Each advertisement is 1 * wi. If the advertisement can be pasted, the position of the advertisement is output (that is, the row number, the position is as close as possible to the left ); otherwise, output-1. analysis: There are only min (n, h) leaf nodes, so do not be scared by 10 ^ 9. Then save the space of each row into the tree, each time an advertisement is pasted, the corresponding space is deleted.

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          #include using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a,b) memset(a,b,sizeof(a))#define MAXN 200010struct node{ int a,b,r;}t[MAXN*4];int h,w,n;void build(int x, int y, int num){ t[num].a = x; t[num].b = y; t[num].r = w; if(x == y) return ; int mid = (x+y)>>1; build(x, mid, num<<1); build(mid+1, y, num<<1|1);}int query(int x, int num){ if(t[num].a == t[num].b) { t[num].r -= x; return t[num].a; } else { int sum1=0, sum2=0; if(x <= t[num<<1].r) sum1 = query(x, num<<1); else if(x <= t[num<<1|1].r) sum2 = query(x, num<<1|1); t[num].r = max(t[num<<1].r, t[num<<1|1].r); return sum1+sum2; }}int main(){ while(scanf("%d%d%d",&h,&w,&n)==3) { if(h > n) h = n; build(1, h, 1); int k; for(int i=1; i<=n; i++) { scanf("%d",&k); if(k <= t[1].r) printf("%d\n",query(k, 1)); else printf("-1\n"); } } return 0;}