Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17122 Accepted Submission(s): 7233
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.
output "-1" for this announcement.
2
4
3
3
3
2
1
3
-1
/*
hdu2795 线段树 贴广告
在一个h*w的广告板上贴广告,每次广告的大小为1*wi
要求每次尽可能贴的高,然后靠左,求每次广告在哪一行
h和w都为10^9,但是总共只有200000个广告,每次广告最多占一行,所以h=min(h,q),
在这样的大小便能使用线段树。每次先检查能够插入,然后在最靠左的位置上加上即可
hhh-2016-02-29 10:42:10
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <vector>
typedef long long ll;
using namespace std;
const int maxn = 200000+5;
ll h,w;
struct node
{
int l,r;
ll Min;
} tree[maxn<<2]; void push_up(int r)
{
int lson = r<<1,rson = (r<<1)|1;
tree[r].Min = min(tree[lson].Min,tree[rson].Min);
}
void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].Min = 0;
if(l == r)
{
return ;
}
int mid = (l+r)>>1;
build(i<<1,l,mid);
build(i<<1|1,mid+1,r);
push_up(i);
}
void push_down(int r)
{ } void Insert(int i,ll k)
{
if(tree[i].l == tree[i].r)
{
tree[i].Min += (ll)k;
printf("%d\n",tree[i].l);
return ;
}
push_down(i);
ll M1 = tree[i<<1].Min;
//ll M2 = tree[i<<1|1].Min;
if(w-M1 >= k) Insert(i<<1,k);
else Insert(i<<1|1,k);
push_up(i);
} int main()
{
ll q;
while(scanf("%I64d%I64d%I64d",&h,&w,&q)!=EOF)
{
h = min(h,q);
build(1,1,h);
ll x;
for(int i =1; i <= q; i++)
{
scanf("%I64d",&x);
if(w-tree[1].Min>=x)
Insert(1,x);
else
printf("-1\n");
}
}
return 0;
}