hdu 2795 Billboard（线段树-单点更新）

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 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

Sample Output

1
2
1
3
-1

题意：有一块尺寸为h*w的矩形长板，要在上面贴1*wi的海报n张，选择贴海报的位置是：尽量高，同一高度，选择尽量靠左的地方。要求输出每张海报的高度位置。

因为海报的高都是1，所以最高的高度就是n,题目要求海报尽可能的高，所以用高来建树最好，每个高都对应一个宽，初始的时候每个高的宽都是w,区间维护最大的宽，当进行更新操作的时候，先判断区间维护的最大宽合不合适，合适的情况下，再缩小区间（也就是尽可能贴高处），再判断，直到区间收缩成一个点，然后更新其宽，再往上更新。

#include <iostream>

#include <stdio.h>

using namespace
std;

#define maxn 200002

int Max[3*maxn];

int h,w,n;

void PushUp(int root)

{

Max[root]=max(Max[root<<1],Max[(root<<1)+1]);

}

void Build(int L,int R,int root)

{

Max[root]=w;//初始都为w

if(L==R)

return;

int m=(L+R)>>1;

Build(L, m, root<<1);

Build(m+1, R, (root<<1)+1);

}

int Query(int x,int L,int R,int root)//查询+更新

{

if(L==R)

{

Max[root]-=x;
//更新宽

return L;

}

int m=(L+R)>>1;

int ans=0;

if(Max[root<<1]>=x)//先往高处贴

ans=Query(x,L,m,root<<1);

else

ans=Query(x,m+1,R,(root<<1)+1);

PushUp(root);
//更新

return ans;

}

int main()

{

int x;

while (scanf("%d%d%d",&h,&w,&n)!=EOF)

{

if(h>n)

h=n;

Build(1,h,1);

for (int i=0; i<n; i++)

{

scanf("%d",&x);

if(Max[1]<x)

{

puts("-1");

continue;

}

printf("%d\n",Query(x,1,h,
1));

}

}

return 0;

}