hdu 2795 Billboard（线段树+单点更新）

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

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13050    Accepted Submission(s): 5651

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：有一块尺寸为h*w的矩形长板，要在上面贴1*wi的海报n张。海报贴的位置要尽量靠左，如果一行能够填满就填满，最后输出的是这张海报贴的高度位置。

 

解题思路:用高度进行建树，每一次比较宽度，在结构体里面定义一个mmax来表示最大的宽度，这个宽度在Updata（）这个函数中不断地更新。一直不断的将以下的最大值反馈到父亲结点。

 

特别注意：1、如果高度是5，但是有2个海报，这样我们就不需要建高度为5的树了，现在看来没有特别大的区别，但是题目中给的数据是1 <= h,w <= 10^9; 1 <= n <= 200,000~~

　　　　　2、这个题目要求输出的是你更新的位置，所以就可以少一个查找的函数。再更新的函数里面，如果这个位置更新了，就可以直接输出这个位置就OK了。

　　　　　3、一定要先比较左结点，在比较右结点。左不行在考虑右。

 

详见代码。

 

#include <iostream>#include <cstdio>using namespace std;struct node{    int l,r;    int mmax;} s[200000*4];void InitTree(int l,int r,int k,int w){    s[k].l=l;    s[k].r=r;    s[k].mmax=w;    if (l==r)        return ;    int mid=(l+r)/2;    InitTree(l,mid,2*k,w);    InitTree(mid+1,r,2*k+1,w);}void UpdataTree(int num,int k)//进行点的更新{    if (s[k].r==s[k].l)    {        printf ("%d\n",s[k].l);        s[k].mmax-=num;        return ;    }    if (num<=s[k*2].mmax)        UpdataTree(num,k*2);    else        UpdataTree(num,k*2+1);    s[k].mmax=s[k*2].mmax>s[k*2+1].mmax?s[k*2].mmax:s[k*2+1].mmax;}int main (){    int h,w,n,a;    while (~scanf("%d%d%d",&h,&w,&n))    {        int hh=h>n?n:h;        InitTree(1,hh,1,w);        for (int i=1; i<=n; i++)        {            scanf("%d",&a);            if (a<=s[1].mmax)                UpdataTree(a,1);            else                printf ("-1\n");        }    }    return 0;}