hdu 2795 Billboard（線段樹-剩餘空間）

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 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.

 

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 5 2 4 3 3 3

題意就是有一塊模板高h寬w，然後有很多海報要往上面貼，每張海報寬wi高1，要貼時都會自頂向下搜索，一旦發現有一行長度足夠就會貼到這一行最左邊還沒被貼到的位置

輸出貼上去的海報所在的行數，沒地方貼輸出-1

用線段樹維護該區間剩餘的最大長度，底部節點則爲每一行的剩餘長度，用區間搜索尋找合適的位置即可。

本題的一個坑點在於線段樹的區間最大值，本來以爲要開到10^9,可是這樣必然會爆掉，其實只需要開到海報張數即可，當然如果高度比海報張數少則開到高度即可。

#include<stdio.h>
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1

const int MAXN=200001;
int lave[MAXN*4];
int w;

void pushup(int rt)
{
	lave[rt]=lave[rt*2]>lave[rt*2+1]?lave[rt*2]:lave[rt*2+1];
}

void build(int l,int r,int rt)
{
	if(l==r)
	{
		lave[rt]=w;
		return ;
	}
	int m=(l+r)/2;
	build(lson);
	build(rson);
	pushup(rt);
}

void paste(int len,int l,int r,int rt)
{
	if(l==r)
	{
		lave[rt]-=len;
		printf("%d\n",l);
		return ;
	}
	int m=(l+r)/2;
	if(len<=lave[rt*2]) paste(len,lson);
	else paste(len,rson);
	pushup(rt);
}

int main()
{
	int h,n,len;
	while(scanf("%d%d%d",&h,&w,&n)!=EOF)
	{
		h=h>n?n:h;
		build(1,h,1);
		while(n--)
		{
			scanf("%d",&len);
			if(len>lave[1]) printf("-1\n");
			else paste(len,1,h,1);			
		}
	}
	return 0;
}