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.

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.

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.

3 5 5
2
4
3
3
3

1
2
1
3
-1

hhanger@zju

HDOJ 2009 Summer Exercise（5）

lcy   |   We have carefully selected several similar problems for you:  1698 1542 1828 1540 2871 

    题意： 有一个h*w的木板，要在上面贴广告，有一个要求就是尽可能的贴在上面，如果不能就尽可能的贴在左面，（这样可以使得贴的广告数目最多吧），每个广告都是1*wi，这就说明每一个广告只能占据一行，可以使用线段树进行维护，每次都进行比较，如果最上面剩下的宽度大于或等于广告的宽就减去广告在木板该行的宽度。如果能想起来这种方法就很简单，如果想不出来都不知道怎么做...

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>

using namespace std;

const int maxn = 200001;

struct node
{
    int l;
    int r;
    int cnt;
}q[maxn<<4];
int h,w,n;

void build(int l,int r,int rt)
{
    q[rt].l = l;
    q[rt].r = r;
    q[rt].cnt = 0;
    if(q[rt].l == q[rt].r)
    {
        q[rt].cnt = w;
        return ;
    }
    int mid = (l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt);
}

int qurry(int k,int l,int r,int rt)
{
    if(q[rt].l == q[rt].r)
    {
        q[rt].cnt -= k;
        return q[rt].l;
    }
    int mid = (l+r)>>1;
    int ans = 0;
    if(q[rt<<1].cnt>=k)
    {
        ans = qurry(k,l,mid,rt<<1);
    }
    else
    {
        ans = qurry(k,mid+1,r,rt<<1|1);
    }
    q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt);
    return ans;
}

int main()
{
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        if(h>n)
        {
            h = n;
        }
        build(1,h,1);
        while(n--)
        {
            int m;
            scanf("%d",&m);
            if(m>q[1].cnt)
            {
                printf("-1\n");
            }
            else
            {
                int ans = qurry(m,1,h,1);
                printf("%d\n",ans);
            }

        }
    }
    return 0;
}