hdu-2795-Billboard(线段树)

                

Billboard

 

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 524333

Sample Output

1213-1

题意：

有个公告板，大小为h*w,要贴n张公告，每个公告的长度是k，高度固定为1，公告放的要尽可能靠上并尽可能靠左，每给出一张公告，要求这个公告在满足要求的情况下放在了第几层。

思路：

因为每个公告的高度一定，可以以公告牌的高度来建树，树从左到右代表了公告牌从上到下，输入公告的长度后，先和w比，如果大于w则直接输出-1.

如果k<w，则进入insert1函数，因为公告先贴到上边，所以优先考虑左子树，每贴一条公告后，就有一个子树剩余的空间就会减少，所以 贴完后要对树进行更新

代码：

 C++ Code 

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#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int h,w,n;
int ans;

struct node
{
    int l,r,n;///左端点，右端点，以及剩余的空白长度，
} a[1000000];

void init(int l,int r,int i,int w)///建树 （左端点，右端点，结构体编号，长度）
{
    a[i].l=l;
    a[i].r=r;
    a[i].n=w;
    if(l!=r)
    {
        ///二分法建树
        int mid=(l+r)/2;
        init(l,mid,2*i,w);///建左树，
        init(mid+1,r,2*i+1,w);///建右树
    }
}

void insert(int i,int x)
{
    if(a[i].l == a[i].r)    ///左端点等于右端点， 到了叶子节点，叶子节点的值既是层数
    {
        a[i].n-=x;      ///该层宽度减少
        ans = a[i].l;
        return ;
    }
    if(x<=a[2*i].n)   ///左子树的宽度可以放下，符合要求搜左子树
        insert(2*i,x);
    else       ///否则右子树
        insert(2*i+1,x);
    a[i].n = max(a[2*i].n,a[2*i+1].n);  ///将左右子树里能放的最大长度存入父亲节点，进行更新
}

int main()
{
    int i,k;
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        if(h>n)///根据题意，因为最多放n个公告，占用的最大高度也只有n
            h = n;
        init(1,h,1,w);///从第一行，间高度为h的树，根节点编号为1，初始化广告板的长度为w;
        ans = -1;
        for(i = 1; i<=n; i++)
        {
            scanf("%d",&k);
            if(a[1].n>=k)    ///如果这个公告没有超出公告板的长度，那么才能放入
            {
                insert(1,k);
                printf("%d\n",ans);
            }
            else
                printf("-1\n");
        }
    }

    return 0;
}