hud 2795

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 524333 

 

Sample Output

 1213-1 

 

题目大意： 在广告牌上贴小广告，告诉你广告牌上的长宽高，然后每个小广告的宽度，每个小广告的高是1，并且两个小广告之间不能覆盖，每次贴的时候总要在最上面，然后求每个广告贴在了哪一行，如果贴不下就输出-1，

分析： 这是一个简单的线段树，（如果想明白的话，想不明白的话会超空间超到哭），建树的时候不能就以h来建树，分析一下，如果有m个广告且m < h;那么m个广告最多贴在前m行，后面的根本没有用，所以建树的时候只需要建m个叶子节点就可以了，所以建树的时候建min（h，n）就可以了；

代码：

#include <iostream>#include <cstdio>#include<cstring>#include<algorithm>#define MAX 200005using namespace std;int tree[MAX << 2];int high,wi,n;void build ( int p,int l, int r ){    if ( l == r ){ tree[p] = wi; return ; }    int mid = ( l + r ) >> 1;    build ( p * 2, l , mid );    build ( p * 2 + 1, mid + 1, r );    tree[p] = max(tree[ 2 * p ] , tree [ 2 * p + 1 ]);}int  change(int p,int l, int r,int post){    if ( l == r ){tree[p] -= post;return l;}    int mid = ( l + r ) >> 1;    int ans = 0;    if ( tree[2*p] >= post )        ans = change(2*p,l,mid,post);    else        ans = change(2*p+1,mid+1,r,post);    tree[p] = max(tree[2*p],tree[2*p+1]);    return ans;}int main(){    while ( cin >> high >> wi >> n )    {        high = min(high,n);        build(1,0,high-1);        for ( int i = 0; i < n; i++)        {            int put;            scanf("%d",&put);            if ( put > tree[1] ) printf("-1\n");            else            {                printf("%d\n",change(1,0,high-1,put)+1);            }        }    }}