hdu 2795 Billboard 曾经无数次TLE

Billboard

Time Limit: 20000/8000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3627Accepted Submission(s): 1720

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
黑板上贴报纸  尽量先贴最上面  在上面的时候 优先贴最左边
问每次 给出报纸的长度  问会贴在哪一行?
思路 一开始想暴力 一看 数据 就哭了 
之后很容易联想到线段树  但是数据为10的9次放
数组开不了那么大啊 想要离散话  
其实不用开那么大 因为最多 因为最多200000层 每层建立一个接点  接点附加信息存储最大的可用长度
一开始一个劲的TLE 就是由于有个大的分支 我全给调用了
#include<stdio.h>
struct haha
{
 int left;
 int right;
 int maxlen;
}node[200000*4];
int len,n,flag,ans,cnt;
void build(int left,int right,int nd)
{
 int mid;
 node[nd].left=left;
 node[nd].right=right;
 node[nd].maxlen=len;
 if(left==right)
 return ;
 mid=(right+left)/2;
 build(left,mid,nd*2);
 build(mid+1,right,nd*2+1);
}
int max(int a,int b)
{
 if(a>b) return a;
 else return b;
}
void update(int x,int nd)
{
 if(flag) return ;
 if(node[nd].maxlen<x) return;
if(node[nd].left==node[nd].right) 
{
node[nd].maxlen-=x;
flag=1;
ans=node[nd].left;
return;
}
if(node[nd*2].maxlen>=x) 
{
update(x,nd*2);
}
else//无数次TLE的原因就在这里少了else 多递归了好多遍
if(node[nd*2+1].maxlen>=x)
{
update(x,nd*2+1);
}
node[nd].maxlen=max(node[nd*2].maxlen,node[nd*2+1].maxlen);
}
int main()
{
 int ti;
while(scanf("%d %d %d",&n,&len,&ti)!=EOF)
{
if(n>200000) n=200010;
build(1,n,1);
int x;
while(ti--)
{
flag=0;
scanf("%d",&x);
ans=-1;
update(x,1);
printf("%d\n",ans);
}
}
return 0;
}