深入理解线段树运作流程(2795)

本文介绍了一个关于公告板布局的问题及其实现算法。该算法利用线段树的数据结构来高效地解决公告板上公告的放置问题,确保每个公告尽可能地放置在顶部且靠左的位置。通过对输入数据的分析和处理,该算法能够输出每个公告最终被放置在哪一行。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17263    Accepted Submission(s): 7294


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

乍一看h范围很大,但是n范围最多20W,所以,多出来的很多行木板是多余的。通过这道题要深刻理解线段树的运作流程:原题要求有相同满足条件时,优先选择靠前的,所以我们在遍历的时候一定要先从左子树开始考虑,tree[1]保存着整个区间的最大值,所以如果最大值也比要插入的小,就可以直接输出-1了。


/*------------------Header Files------------------*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctype.h>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <vector>
#include <limits.h>
using namespace std;
/*------------------Definitions-------------------*/
#define LL long long
#define PI acos(-1.0)
#define INF 0x3F3F3F3F
#define MOD 1E9+7
#define MAX 500050
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
/*---------------------Work-----------------------*/
int h,w,n,tree[200050<<2];
void pushup(int rt)
{
	tree[rt]=max(tree[rt<<1],tree[rt<<1|1]);
}
void build(int rt,int l,int r)
{
	tree[rt]=w;
	if(l==r) return ;
	int m=(l+r)>>1;
	build(lson);
	build(rson);
}
int query(int rt,int l,int r,int x)
{
	if(l==r) //肯定要遍历到一个叶子结点为止的
	{
		tree[rt]-=x;
		return l;
	}
	int m=(l+r)>>1,ans;
	if(tree[rt<<1]>=x) ans=query(lson,x); //左子树优先考虑
	else ans=query(rson,x);
	pushup(rt);
	return ans;
}
void work()
{
	while(scanf("%d%d%d",&h,&w,&n)==3)
	{
		if(h>n) h=n; //这一步很关键
		build(1,1,h);
		int ans;
		while(n--)
		{
			scanf("%d",&ans);
			if(tree[1]<ans) printf("-1\n");
			else printf("%d\n",query(1,1,h,ans));
		}
	}
}
/*------------------Main Function------------------*/
int main()
{
	//freopen("test.txt","r",stdin);
	//freopen("cowtour.out","w",stdout);
	//freopen("cowtour.in","r",stdin);
	work();
	return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值