HDU 2795 Billboard

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18408    Accepted Submission(s): 7716

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

hhanger@zju

 

Source

HDOJ 2009 Summer Exercise（5）

 

 

 

解析：线段树，单点更新。以高度为基准来建树，乍一看，h可能达到10^9，不可能开到10^9的四倍大的数组。但是，只进行n次操作，在最坏的情况下，每一次操作，公告都单独占据一行，n个公告占据了前n行的位置，后面的位置根本用不到。所以h取h和n中二者的较小值即可。

 

 

 

#include <cstdio>
#include <algorithm>
using namespace std;

int val[200005<<2]; //记录该行剩余的宽度
int h, w, n;

void pushUp(int v)
{
    val[v] = max(val[v<<1], val[v<<1|1]);
}

void build(int v, int l, int r)
{
    if(l == r){
        val[v] = w;
        return;
    }
    int mid = (l+r)>>1;
    build(v<<1, l, mid);
    build(v<<1|1, mid+1, r);
    pushUp(v);
}

int query(int v, int l, int r, int c)
{
    if(val[v] < c)
        return -1;
    if(l == r){
        val[v] -= c;
        return l;
    }
    int mid = (l+r)>>1;
    int ans = query(v<<1, l, mid, c);  //先查找左子树
    if(ans == -1)   //如果在左子树中找不到合适的位置，则在右子树中查找
        ans = query(v<<1|1, mid+1, r, c);
    pushUp(v);
    return ans;
}

int main()
{
    while(~scanf("%d%d%d", &h, &w, &n)){
        if(h > n)
            h = n;
        build(1, 1, h);
        int wi;
        while(n--){
            scanf("%d", &wi);
            printf("%d
", query(1, 1, h, wi));
        }
    }
    return 0;
}