http://acm.hdu.edu.cn/showproblem.php?pid=2795

Billboard

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

#include <cstdio>
#include <cstring>
using namespace std;
#define N 200005
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
struct node
{
    int value;
}tree[N<<2];
/*
线段树
记得要用h和n的较小值来建树
思路比较难想：每个区间的value装的是该区间还能放下的最大长度
然后与输入进来的len进行比较
题意要求优先放左边，所以如果左儿子放得下优先放左儿子，不然看右儿子
然后要PushUp更新父节点信息
如果根节点即第一个节点放不下，即整棵树必定没有一个区间可以放得下该广告
*/
void Build(int rt,int l,int r,int k)
{
    tree[rt].value=k;
    if(l==r) return ;
    int m=(l+r)>>1;
    Build(lson,k);
    Build(rson,k);
}

int Query(int rt,int l,int r,int len)
{
    int ans;
    if(l==r){
        tree[rt].value-=len;
        return l;
    }
    int m=(l+r)>>1;
//如果左边能放下，优先放左边，不然看右边能不能放下
    if(tree[rt<<1].value>=len) ans=Query(lson,len);
    else ans=Query(rson,len);
//    tree[rt].value=max(tree[rt<<1].value,tree[rt<<1|1].value);
//PushUp
    if(tree[rt<<1].value>=tree[rt<<1|1].value)
        tree[rt].value=tree[rt<<1].value;
    else tree[rt].value=tree[rt<<1|1].value;
    return ans;
}

int main()
{
    int h,w,n;
    while(~scanf("%d%d%d",&h,&w,&n)){
        if(h>n) h=n;
        Build(1,1,h,w);
        while(n--){
            int len;
            scanf("%d",&len);
            if(tree[1].value<len) printf("-1\n");
            else printf("%d\n",Query(1,1,h,len));
        }
    }
    return 0;
}