hdu2795 Billboard 线段树应用

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

Hint

题意

题解：

想办法转化成线段树模型
从高度入手 建立单位区间
查询长度到最底部的区间即单位区间 这个区间的端点值即对应可以插入的高度位置
判断查询长度满足的区间即可 如果当前区间内的最大长度还小于查询长度或者已经找到满足条件的位置就跳出

AC代码

#include<stdio.h>#include <math.h>#include <cstring>#include<algorithm>using namespace std;const int MAXN = 2e5+5;struct node{    int mx;}tree[MAXN<<2];int h,w,n;void Buildtree(int o,int l,int r){    if (l  == r){        tree[o].mx = w;        return ;    }    int mid = (l+r)>>1;    Buildtree(o<<1,l,mid);    Buildtree(o<<1|1,mid+1,r);    tree[o].mx = max(tree[o<<1].mx,tree[o<<1|1].mx);}int flag;void  Srch(int o,int l,int r,int x){    if (tree[o].mx<x||flag) return ;    if (l == r){        if (tree[o].mx>=x){            printf("%d\n",l);            tree[o].mx -= x;            flag = 1;            return ;        }    }    int mid = (l+r)>>1;    if (x <= tree[o<<1].mx) Srch(o<<1,l,mid,x);    else Srch(o<<1|1,mid+1,r,x);    tree[o].mx = max(tree[o<<1].mx,tree[o<<1|1].mx);}int main(){    while (scanf("%d%d%d",&h,&w,&n)!=EOF){            memset(tree,0,sizeof(tree));            int x;            if (h > MAXN) h = MAXN;            Buildtree(1,1,h);            while (n--){                scanf("%d",&x);                flag = 0;                Srch(1,1,h,x);                if (!flag) printf("-1\n");            }    }    return 0;}