hdu 2795 Billboard （线段树）

最新推荐文章于 2019-07-20 21:43:37 发布

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本文介绍了一个基于线段树的数据结构算法，用于解决公告牌布局问题。该算法通过维护每个区间的最大剩余宽度，实现了在公告牌上高效地放置不同宽度的公告。当新的公告无法放置时，算法会返回-1，表示没有合适的位置。此算法适用于处理大量公告的快速布局，确保公告尽可能地靠上靠左。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2795
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
题目大意：一个h*w的公告牌，要在其上贴公告。
输入的是1*wi的w值，这些是公告的尺寸
接下来要满足的条件有：1、尽量往上，同一高度尽量靠左。2、求第n个广告所在的行数。3、没有合适
的位置贴了则输出-1。

用线段树求出区间最大值，在结构体中定义一个MAX来确定当前区间还剩余的最大长度，然后与输入进
来的长度比较。
#pragma GCC optimize(2)
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<string>
#include<queue>
using namespace std;
const int maxn = 2e5 + 10;
const int inf = 0x3f3f3f3f;
typedef long long ll;
struct node
{
	int l, r;
	int Max;
}edge[maxn * 4];
int h, w, n;
void buildtree(int l, int r, int root, int w)
{
	edge[root].l = l;
	edge[root].r = r;
	edge[root].Max = w;
	if (l == r)
	{
		return;
	}
	int mid = (l + r) / 2;
	buildtree(l, mid, root * 2, w);
	buildtree(mid + 1, r, root * 2 + 1, w);
}
void update(int x, int root)
{
	if (edge[root].l == edge[root].r)
	{
		printf("%d\n", edge[root].l);
		edge[root].Max -= x;
		return;
	}
	if (x <= edge[root * 2].Max)
	{
		update(x, root * 2);
	}
	else
	{
		update(x, root * 2 + 1);
	}
	edge[root].Max = max(edge[root * 2].Max, edge[root * 2 + 1].Max);
}
int main()
{
	freopen("C://input.txt", "r", stdin);
	while (scanf("%d%d%d", &h, &w, &n) != EOF)
	{
		if (h > n)
		{
			h = n;
		}
		buildtree(1, h, 1, w);
		for (int i = 1; i <= n; i++)
		{
			int p;
			scanf("%d", &p);
			if (edge[1].Max >= p)
			{
				update(p, 1);
			}
			else
			{
				printf("-1\n");
			}
		}
	}
	return 0;
}