VJudge--Billboard

题目 - Billboard

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：

​ 第一行输入的前两个数是给的布告栏的高和宽，第三个数是有几个通知，每个通知的的高都是单位长度，下面几行是每个通知的宽要求输出每个通知分别贴在第几行。优先选择最上边和最左边的地方贴。

解题：

​ 其实一开始是没有想到用线段树来解题，因为感觉没办法用线段树？我的常规思路是把宽度来当成一个区间，实际上这个题目是把高度当区间来建树，感觉还是挺有收获的，写出来之后一直RE，最后发现我把l，r写成了r,l。这种手误情况还是要尽量避免哇。

AC代码：

#include<iostream>
#include<string.h>
#include<cstdio>
#include<algorithm>

using namespace std;

const int MAXN = 200010;

int h, w;
int tree[MAXN << 2];

void push_up(int u) {
	tree[u] = max(tree[u << 1], tree[u << 1 | 1]);
}

void BuildTree(int l, int r, int u) {
	if(l == r) {
		tree[u] = w;
		return;
	}
	int mid = (l + r) >> 1;
	BuildTree(l, mid, u << 1);
	BuildTree(mid + 1, r, u << 1 | 1);
	push_up(u);
}

void Update(int x, int l, int r, int u, int ans) {
	if( l == r) {
		tree[u] -= x;
		return;
	}
	int mid = (l + r) >> 1;
	if(ans <= mid) Update(x, l, mid, u << 1, ans);
	else Update(x, mid + 1, r, u << 1 | 1, ans);
	push_up(u);
}

int Query(int x, int l, int r, int u) {
	if(l == r) return l;
	int mid = (l + r) >> 1;
	int ans = 0;
	if(tree[u << 1] >= x) ans = Query(x, l, mid, u << 1);
	else if (tree[u << 1 | 1] >= x) ans = Query(x, mid + 1, r, u << 1 | 1);
	return ans;
}

int main() {
	int n;
	while(~scanf("%d%d%d", &h, &w, &n)) {
		if(h > n) h = n;
		BuildTree(1, h, 1);
		for(int i = 1; i <= n; i++) {
			int x;
			scanf("%d", &x);
			if(tree[1] < x) {
				printf("-1\n");
				continue;
			}
			int ans = Query(x, 1, h, 1);
			printf("%d\n", ans);
			Update(x, 1, h, 1, ans);
		}
	}
	return 0;
}