UVA - 10043 Chainsaw Massacre

Description

 Problem E: Chainsaw Massacre 

Background

As every year the Canadian Lumberjack Society has just held its annual woodcutting competition and the national forests between Montreal and Vancouver are devastated. Now for the social part!In order to lay out an adequate dance floor for the evening partythe
organizing committee is looking for a large rectangular area without trees. Naturally, all lumberjacks are already drunk and nobody wants to take the risk of having any of them operate a chainsaw.

The Problem

The organizing committee has asked you to find the largest yet freerectangle which could serve as the dance floor. The area inwhich you should search is also rectangular and the dance floor mustbe entirely located in that area.Its sides should be parallel to
the borders of the area.It is allowed that the dance floor is located at the borders of the areaand also that trees grow on the borders of the dance floor.What is the maximum size of the dance floor?

The Input

The first line of the input specifies the number of scenarios. For each scenario, the first line provides the length
l and widthw of the area in meters (,both integers). Each ofthe following lines describes either a single
tree, or a line of treesaccording to one of the following formats:

• 1 x y, where the ``one'' characterizes a single tree, and x and
  y provide its coordinates in meters with respect to the upper leftcorner.
• k x y dx dy, where k>1 provides the number of trees in a line withcoordinates
  .
• 0 denotes the end of the scenario.

The coordinates x, y, dx, and dy are given as integers. It is guaranteed that all the trees are situated in the area, i.e. have coordinatesin
.There will be at most 1000 trees.

The Output

For each scenario print a line containing the maximum size of the dance floor measured in square meters.

Sample Input

2
2 3
0
10 10
2 1 1 8 0
2 1 9 8 0
0

Sample Output

6
80

题意：平面上有n棵树。找出一个内部没有树的，面积最大的矩形

思路：以y坐标排序然后扫描。每次先扫到一棵树就能够知道它与上一棵树之间的距离，然后更新统计每一个x坐标的最左边和最右边，每次都计算一次

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <map>
#include <vector>
using namespace std;
const int maxn = 10010;

int h[maxn], l[maxn], r[maxn];
int n, m, ans;
map<int, vector<int> > tree;

void check() {
	for (int i = 0, j = n; i <= n; i++, j--) {
		for (l[i] = i; l[i] > 0 && h[l[i]-1] >= h[i]; )
			l[i] = l[l[i]-1];
		for (r[j] = j; r[j] < n && h[r[j]+1] >= h[j]; )
			r[j] = r[r[j]+1];
	}
}

void cal() {
	for (int i = 0; i <= n; i++) {
		int tmp = h[i] * (r[i] - l[i] + 2);
		ans = max(ans, tmp);
	}
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		tree.clear();
		scanf("%d%d", &n, &m);
		int op, x, y, dx, dy;
		while (1) {
			scanf("%d", &op);
			if (op == 0)
				break;
			else if (op == 1) {
				scanf("%d%d", &x, &y);
				tree[y].push_back(x);
			}
			else {
				scanf("%d%d%d%d", &x, &y, &dx, &dy);
				for (int i = 0; i < op; i++) {
					tree[y].push_back(x);
					y += dy, x += dx;
				}
			}
		}
		tree[m];
		ans = max(n, m);
		int last = 0;
		memset(h, 0, sizeof(h));
		map<int, vector<int> >::iterator it;
		for (it = tree.begin(); it != tree.end(); it++) {
			int d = it->first - last;
			last += d;
			for (int i = 1; i < n; i++)
				h[i] += d;
			check();
			cal();
			vector<int> tmp = it->second;
			for (int i = 0; i < tmp.size(); i++)
				h[tmp[i]] = 0;
		}
		printf("%d\n", ans);
	}
	return 0;
}