HDU【2795】Billboard

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16934    Accepted Submission(s): 7159

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    

      /*     这道题建立一个最大长度为n的线段树，这里将每个根节点的值都设为w，     即最大宽度，每个父节点为所有子节点中w的最大值，用于判断下面是否存在     可以放下宽度为x的公告，这里同时体现出了线段树提高查找效率的地方。找到     的话就可以将叶子节点的值减去板报的宽度，此时就是剩下可以剩下能放的     公告宽度了。 */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; 
    
    
    
    
    
 
    
    
    
    
    

      const int maxn = 200005; int h,w,n,x; int tree[4*maxn]; 
    
    
    
    
    
 
    
    
    
    
    

      void Build(int root,int L,int R) {     tree[root] = w;     if(L == R)         return;     int M = (L+R)/2;     Build(2*root , L , M);     Build(2*root+1 , M+1 , R); } 
    
    
    
    
    
 
    
    
    
    
    

      int Query(int root,int L,int R) {     if(L == R)     {         tree[root] -= x;         return L;     }     int M = (L+R)/2,ret;     if(x <= tree[root*2])         ret = Query(root*2, L, M);     else         ret = Query(root*2+1, M+1, R);     tree[root] = max(tree[root*2],tree[root*2+1]);     return ret; } 
    
    
    
    
    
 
    
    
    
    
    

      int main() {     while(scanf("%d%d%d",&h,&w,&n) != EOF)     {         int y = n;         if(h > n)  h = n;         Build(1,1,h);         while(y--)         {             scanf("%d",&x);             if(x > tree[1]) printf("-1\n"); //这里通过判断根节点的值来判断这块板报是否放的下             else                 printf("%d\n",Query(1,1,h));         }     }     return 0; }