hdu2795 线段树 贴广告

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17122    Accepted Submission(s): 7233

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

/*
hdu2795 线段树 贴广告
在一个h*w的广告板上贴广告，每次广告的大小为1*wi
要求每次尽可能贴的高，然后靠左，求每次广告在哪一行
h和w都为10^9，但是总共只有200000个广告，每次广告最多占一行，所以h=min(h,q)，
在这样的大小便能使用线段树。每次先检查能够插入，然后在最靠左的位置上加上即可
hhh-2016-02-29 10:42:10
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <vector>
typedef long long ll;
using namespace std;
const int maxn = 200000+5;
ll h,w;
struct node
{
    int l,r;
    ll Min;
} tree[maxn<<2];

void push_up(int r)
{
    int lson = r<<1,rson = (r<<1)|1;
    tree[r].Min = min(tree[lson].Min,tree[rson].Min);
}
void build(int i,int l,int r)
{
    tree[i].l = l,tree[i].r = r;
    tree[i].Min  = 0;
    if(l == r)
    {
        return ;
    }
    int mid = (l+r)>>1;
    build(i<<1,l,mid);
    build(i<<1|1,mid+1,r);
    push_up(i);
}
void push_down(int r)
{

}

void Insert(int i,ll k)
{
    if(tree[i].l == tree[i].r)
    {
        tree[i].Min += (ll)k;
        printf("%d\n",tree[i].l);
        return ;
    }
    push_down(i);
    ll M1 = tree[i<<1].Min;
    //ll M2 = tree[i<<1|1].Min;
    if(w-M1 >= k) Insert(i<<1,k);
    else Insert(i<<1|1,k);
    push_up(i);
}

int main()
{
    ll q;
    while(scanf("%I64d%I64d%I64d",&h,&w,&q)!=EOF)
    {
        h = min(h,q);
        build(1,1,h);
        ll x;
        for(int i =1; i <= q; i++)
        {
            scanf("%I64d",&x);
            if(w-tree[1].Min>=x)
                Insert(1,x);
            else
                printf("-1\n");
        }
    }
    return 0;
}