HDU 2795 - Billboard(线段树)

Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20325 Accepted Submission(s): 8442

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

题意：
给出一个矩形，长度为h，宽度为w。
现在给你n个宽度为wi，长度为1的矩形，要放在大矩形中。
策略是从高往低，从左往右放置。
问每一个给出的小矩形放置的层数，如果不能放置，就输出1.

解题思路：
用线段树的思想，每一个节点维护的是区间中叶节点剩余的最大空间，叶节点就是每一层的剩余空间。
因为最多放置n层，那么h就根据n来进行更新。

AC代码：

#include<bits/stdc++.h>#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;const int maxn = 2e5;int stree[maxn<<2];void build(int l,int r,int rt,int w){    stree[rt] = w;    if(l == r)  return ;    int mid = (l+r)>>1;    build(lson,w);    build(rson,w);}void PushUp(int rt) {stree[rt] = max(stree[rt<<1],stree[rt<<1|1]);}int query(int l,int r,int rt,int wi){    if(l == r)    {        stree[rt] -= wi;        return l;    }    int mid = (l+r)>>1;    int res = wi<=stree[rt<<1]?query(lson,wi):query(rson,wi);    PushUp(rt);    return res;}int main(){    int h,w,n;    while(~scanf("%d%d%d",&h,&w,&n))    {        if(h > n)   h = n;        build(1,h,1,w);        while(n--)        {            int wi;            scanf("%d",&wi);            if(wi > stree[1])   printf("-1\n");            else                printf("%d\n",query(1,h,1,wi));        }    }    return 0;}