Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4292 Accepted Submission(s): 2003
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.
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.
be put on the billboard, output "-1" for this announcement.
3 5 5 2 4 3 3 3
1 2 1 3 -1
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define N 200100 using namespace std; class node { public: int l,r,ma; }root[4*N]; int n,q; int M; void build(int t,int l,int r,int m) { root[t].l=l;root[t].r=r; if(root[t].l==root[t].r)root[t].ma=m; else { build(t<<1,l,(l+r)/2,m);build(t<<1|1,(l+r)/2+1,r,m); root[t].ma=max(root[t<<1].ma,root[t<<1|1].ma); } } void update(int t,int pos) { int s; if(root[t].l==root[t].r&&root[t].ma>=pos){M=root[t].l;root[t].ma-=pos;return;} if(root[t].ma<pos){return;} else { if(root[t<<1].ma>=pos)update(t<<1,pos); else update(t<<1|1,pos); root[t].ma=max(root[t<<1].ma,root[t<<1|1].ma); } } int main() { int m,a; while(scanf("%d%d%d",&n,&m,&q)!=EOF) { if(n>N) n=N-10; build(1,1,n,m); for(int i=0;i<q;i++) { M=-1; scanf("%d",&a); update(1,a); printf("%d\n",M); } } }