Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

Author

[email protected]

Source

HDOJ 2009 Summer Exercise（5）

十一点半了~  哎呀 ！~   和sb疯一块儿刷题~。刚刷了一个线段树~ 好无聊……   明天 有是杭电多校联合 加油！

这个题目猛一看真的看不出来是线段数，还是做的线段树题目少啊。公示牌儿的高度可以看作线段树根节点的长度，宽度可以看作，每个节点的值，那么结构体元素需要一个保存每个结点中可插入位置的最大值，由于update（）函数性质恰好满足优先往左下角查询插入位置。一旦找到插入位置递归更新此点信息，并返回结果。注意广告牌儿的高度

是一个幌子而已。然后就是代码了：

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
int h,x,n;
struct Tree
{
    int left,right,Max;
} tree[200000<<2];

int create(int root ,int left,int right)
{
    tree[root].left=left;
    tree[root].right=right;
    if(left==right)
    {
        return   tree[root].Max=n;
    }
    int a,b,mid=(left+right)>>1;
    a=create(root<<1,left,mid);
    b=create(root<<1|1,mid+1,right);
    return tree[root].Max=max(a,b);
}

int update(int root,int w)
{
    int res;
    if(tree[root].left==tree[root].right)
    {
        tree[root].Max-=w;
        return tree[root].left;
    }
    if(tree[root<<1].Max>=w)
        res=  update(root<<1,w);
    else
        res= update(root<<1|1,w);
    tree[root].Max=max(tree[root<<1].Max,tree[root<<1|1].Max);
    return res;
}

int main()
{
    while(cin>>h>>n>>x)
    {
        if(h>x)
            h=x;
        create(1,1,h);
        for(int i=1; i<=x; i++)
        {
            int w;
            scanf("%d",&w);
            if(tree[1].Max<w)
                printf("-1\n");
            else
                printf("%d\n",update(1,w));
        }
    }
}

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