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h×w大的公告板。有n条公告要写入，每条公告高度都是1，宽度是wi，每次从最上最左的空位写，假设有空位输出第几行。假设没有足够空位输出-1。
分析：
用高度来构建线段树，记录left行到right行中的最大留白空间，然后每贴一个广告，就在线段树中插，如果左子树的空位长度大于当前要贴的广告长度，就先查找左子树，（因为题目要求贴最左上方），当找到贴的位置，则该行的新的剩余长度就在原值上减去当前长度。并更新整棵线段树。
代码：
#include <iostream>
#include <stdio.h>
#define lchild left,mid,root<<1
#define rchild mid+1,right,root<<1|1
using namespace std;
const int maxn=1e8;
int Max[maxn<<2]; ///用来存放某些行区间所能存放的最大值
int height,width,n;
///构建线段树
void push_up(int root)
{
Max[root] = max(Max[root<<1],Max[root<<1|1]);
}
void build(int left,int right,int root)
{
Max[root] = width; ///初始，每个区间所剩余的最大长度都为width
if(left == right) return;
int mid = (left+right)>>1;
///递归构建左右子树
build(lchild);
build(rchild);
}
///查询并插入
int query(int w,int left,int right,int root)
{
if(left == right) ///则将该板插入到这一行,这一节点所剩余的宽度减去w
{
Max[root] -= w;
return left; ///返回木板所插入的行
}
int mid = (left+right)>>1;
int ans=0;
if(w<=Max[root<<1]) ///如果广告的宽度小于左区间的最大宽度，题目要求往左区间靠
ans = query(w,lchild);
else
ans = query(w,rchild);
push_up(root);
return ans;
}
int main()
{
while(~scanf("%d%d%d",&height,&width,&n))
{
if(n<height) ///所要贴的广告数小于高度，则线段树按n建，否则按height建立
{
height = n;
}
build(1,height,1); ///构建线段树
int w;
while(n--)
{
scanf("%d",&w);
if(Max[1]<w)
printf("-1
");
else
{
int ans = query(w,1,height,1);
printf("%d
",ans);
}
}
}
return 0;
}