线段数小结之（一）hdu 2795Billboard ！！

Billboard

Time Limit : 20000/8000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 3

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

题目的时间limit是8秒  反而有种不祥的预感！题目意思很容易读懂，但真的理解就在与巧妙！！题目的意思是在布告板上贴报纸，从高出优先，

在同高的基础上左边优先，而且报纸很规范，单位长度高！！！！！！得看到这点！

    题意说h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) 

	h这么大，线段树不好建啊！！但是看看N，不大！！那H这么大也没用！ 想一下！h比n大 我贴的海报一行一张也贴不完！故因此

只需建立seg_tree[3*200000]的长度足以！！

不多说！ 贴代码 自己理解！

#include<iostream>
using namespace std;

const __int64 N=200000;
struct seg_tree
{
 int left,right;
 int size;
}t[4*N];
inline int MID(int a,int b)
{
 return (a+b)/2;
}
inline int MAX(int a,int b)
{
 return a>b?a:b;
}
int n,h,w,flag;

void make_tree(int c,int l,int r)
{
 t[c].left = l;
 t[c].right = r;
 t[c].size = w;
 if(l == r)
 {
  return;
 }
 int mid = MID(l,r);
 make_tree (2*c,l,mid );
 make_tree(2*c+1,mid+1,r);
}

void find_ans(int c,int val)
{
 if(val > t[c].size )
 {
  return;
 }
 if(t[c].left==t[c].right)
 {
  t[c].size -= val;
  flag = t[c].left;
  return;
 }
 if(t[2*c].size >= val)
  find_ans(c*2,val);
 else
  find_ans(2*c+1,val);
 t[c].size = MAX(t[c*2].size,t[c*2+1].size);
}

int main()
{
 int i,temp;
 while(cin>>h>>w>>n)
 {
  h = h<n?h:n;
  make_tree(1,1,h);
  for(i=1;i<=n;i++)
  {
   flag = 0;
   scanf("%d",&temp);
   find_ans(1,temp);
   if(flag!=0)
    printf("%d/n",flag);
   else
    printf("-1/n");
  }
 }
 return 0;
}