HDU 2795 Billboard （线段树+想法）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2795

题面：

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16167    Accepted Submission(s): 6837

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
   
   
   
   

题目大意：

    给一个H*M的板，每次贴1*w的海报，要求贴在能贴的行数最小的行的最左边，且不能覆盖，若不能贴，则输出-1，否则输出该行行号。

解题：

    操作是很简单的线段树，但不得不说，这题考的是想法，我自己是想不到用线段树去维护区间行的最大值。线段树熟悉之后，更多得考的是想法和如何灵活地应用线段树。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define maxn 200010
using namespace std;
int maxw[maxn<<2];
int h,w,n,ans,tw;
int max(int a,int b)
{
    return a>b?a:b;
}
//建树 
void build(int le,int ri,int u)
{
   //初值都赋为w 
   maxw[u]=w;
   if(le==ri)return;
   int mid=(le+ri)>>1;
   build(le,mid,u<<1);
   build(mid+1,ri,(u<<1)+1);
}

int query(int le,int ri,int u,int val)
{
    int res;
    if(le==ri)
    {
    	//更新 
        maxw[u]-=val;
        return le;
    }
    //如果左边可以放下，就放左边，即满足尽量放上边 
    if(maxw[u<<1]>=val)
       res=query(le,(le+ri)>>1,u<<1,val);
    //左边放不下，放右边 
    else
       res=query(((le+ri)>>1)+1,ri,(u<<1)+1,val);
    //更新当前节点最大值 
    maxw[u]=max(maxw[u<<1],maxw[(u<<1)+1]);
    return res;
}
int main()
{
    while(~scanf("%d%d%d",&h,&w,&n))
    {
        if(h>n)
          h=n;
        build(1,h,1);
        for(int i=0;i<n;i++)
        {
           scanf("%d",&tw);
           //如果整个区间都放不下了，那么肯定放不下了 
           if(maxw[1]<tw)
             printf("-1\n");
           else
           {
             ans=query(1,h,1,tw);
             printf("%d\n",ans);
           }
        }
    }
	return 0;
}