poj 1389Area of Simple Polygons(矩形面积并)

Area of Simple Polygons

Description


There are N, 1 <= N <= 1,000 rectangles in the 2-D xy-plane. The four sides of a rectangle are horizontal or vertical line segments. Rectangles are defined by their lower-left and upper-right corner points. Each corner point is a pair of two nonnegative integers in the range of 0 through 50,000 indicating its x and y coordinates. 


Assume that the contour of their union is defi ned by a set S of segments. We can use a subset of S to construct simple polygon(s). Please report the total area of the polygon(s) constructed by the subset of S. The area should be as large as possible. In a 2-D xy-plane, a polygon is defined by a finite set of segments such that every segment extreme (or endpoint) is shared by exactly two edges and no subsets of edges has the same property. The segments are edges and their extremes are the vertices of the polygon. A polygon is simple if there is no pair of nonconsecutive edges sharing a point. 


Example: Consider the following three rectangles: 


rectangle 1: < (0, 0) (4, 4) >, 


rectangle 2: < (1, 1) (5, 2) >, 


rectangle 3: < (1, 1) (2, 5) >. 


The total area of all simple polygons constructed by these rectangles is 18. 
Input


The input consists of multiple test cases. A line of 4 -1's separates each test case. An extra line of 4 -1's marks the end of the input. In each test case, the rectangles are given one by one in a line. In each line for a rectangle, 4 non-negative integers are given. The first two are the x and y coordinates of the lower-left corner. The next two are the x and y coordinates of the upper-right corner.
Output


For each test case, output the total area of all simple polygons in a line. 
Sample Input


0 0 4 4
1 1 5 2
1 1 2 5
-1 -1 -1 -1
0 0 2 2
1 1 3 3
2 2 4 4
-1 -1 -1 -1
-1 -1 -1 -1　 
Sample Output


18

10 

solution：

首先将每个矩形用左右两边表示并标记每个线段是左边的边还是右边的边，然后对这些线段按x坐标升序排序，用一根线从左往右扫描这些线段，那么相邻两根扫描线之间围成的矩形面积之和即为面积并，那么我们只需要知道每个扫描线上被线段覆盖的长度，将其乘以相邻两根扫描线之间的距离即为扫描线扫过的面积，因此我们对y轴建一棵线段树（对所有线段的纵坐标离散化后建树），那么每条线段就可以用线段树中的一个区间来表示，每次扫到左边的边就在线段树中覆盖这条边对应的区间，扫到右边的边就在线段树中取消对这条边的覆盖，每扫过一根线段就维护一下扫描线被覆盖的长度即可 首先将每个矩形用左右两边表示并标记每个线段是左边的边还是右边的边，然后对这些线段按x坐标升序排序，用一根线从左往右扫描这些线段，那么相邻两根扫描线之间围成的矩形面积之和即为面积并，那么我们只需要知道每个扫描线上被线段覆盖的长度，将其乘以相邻两根扫描线之间的距离即为扫描线扫过的面积，因此我们对y轴建一棵线段树（对所有线段的纵坐标离散化后建树），那么每条线段就可以用线段树中的一个区间来表示，每次扫到左边的边就在线段树中覆盖这条边对应的区间，扫到右边的边就在线段树中取消对这条边的覆盖，每扫过一根线段就维护一下扫描线被覆盖的长度即可 

#include
#include
#include
#include
using namespace std;
#define ls (t<<1)
#define rs (t<<1)|1
#define eps 1e-8
#define f(a)for(int i=0;i b.flag;
	else return a.x < b.x;
}
void build(int l, int r, int t)
{
	e[t].l = l; e[t].r = r; e[t].num = 0; e[t].len = 0;
	if (l + 1 == r)return;
	int mid = (l + r) >> 1;
	build(l, mid, ls);
	build(mid, r, rs);
}
bool cc(double x, double y)
{
	return fabs(x - y) < eps;
}
void update_len(int t)
{
	if (e[t].num)e[t].len = y[e[t].r] - y[e[t].l];
	else if (e[t].l + 1 == e[t].r)e[t].len = 0;
	else e[t].len = e[ls].len + e[rs].len;
}
void update(double l, double r, int t, int z)
{
	if (cc(y[e[t].l], l) && cc(y[e[t].r], r))e[t].num += z;
	if (e[t].l + 1 < e[t].r)
	{
		int mid = (e[t].l + e[t].r) >> 1;
		if (r <= y[mid] + eps)update(l, r, ls, z);
		else if (l + eps >= y[mid])update(l, r, rs, z);
		else {
			update(l, y[mid], ls, z);
			update(y[mid], r, rs, z);
		}
	}
	update_len(t);
}
int main()
{
	double xx, yy, x2, y2;
	while (scanf("%lf%lf%lf%lf", &xx, &yy, &x2, &y2))
	{
		if (xx + yy + x2 + y2 == -4.0)break;
		n = 0; no = 0;
		while (1)
		{
			add(xx, yy, y2, yy, 1);
			add(x2, yy, y2, y2, 0); n++;
			scanf("%lf%lf%lf%lf", &xx, &yy, &x2, &y2);
			if (xx + yy + x2 + y2 == -4.0)break;
		}
		sort(y, y + no);
		sort(line, line + no, cmp);
		int res = unique(y, y + no) - y;
		build(0, res - 1, 1);
		double ans = 0, len = 0;
		f(no)
		{
			if (line[i].flag)update(line[i].y1, line[i].y2, 1, 1);
			else update(line[i].y1, line[i].y2, 1, -1);
			if (i)ans += len*(line[i].x - line[i - 1].x);
			len = e[1].len;
		}
		printf("%.lf\n", ans);
	}
}