每个方向上最多200个坐标，复杂度最多也就200 * 200

贴个程序：

#include <stdio.h>
#include <vector>
#include <algorithm>

const int L = 100;
double x1[L], x2[L], y1[L], y2[L];

class YSegment {
public:
	double y_begin, y_end;
	bool available;
	bool operator < (const YSegment &rhs) const {
		return this->y_begin < rhs.y_begin;
	}
	YSegment(double y1, double y2, bool a) {
		y_begin = y1;
		y_end = y2;
		available = a;
	}
};

void processCase(int caze, int n) {
	std::vector<double> x_coor(2 * L);
	for (int i = 0; i < n; i++) {
		x_coor.push_back(x1[i]);
		x_coor.push_back(x2[i]);
	}
	std::sort(x_coor.begin(), x_coor.end());

	double area = 0.0;
	for (int j = 0; j < x_coor.size() - 1; j++) {
		double x_left = x_coor[j], x_right = x_coor[j + 1];
		if (x_left == x_right) continue;

		// Sort Y segments.
		std::vector<YSegment> y_segs;
		for (int i = 0; i < n; i++) {
			if (x_left >= x1[i] && x_right <= x2[i]) {
				YSegment y_seg(y1[i], y2[i], true);
				y_segs.push_back(y_seg);
			}
		}
		std::sort(y_segs.begin(), y_segs.end());

		// Combine Y segments
		for (int i = 1; i < y_segs.size(); i++) {
			// Find the previous available segment.
			int p = i - 1;
			while (!y_segs[p].available) p--;
			if (y_segs[i].y_begin <= y_segs[p].y_end) {
				y_segs[i].available = false;
				y_segs[p].y_end = std::max(y_segs[p].y_end, y_segs[i].y_end);
			}
		}

		// Accumulate Y segments
		double y_length = 0.0;
		for (int i = 0; i < y_segs.size(); i++) {
			if (y_segs[i].available) {
				y_length += y_segs[i].y_end - y_segs[i].y_begin;
			}
		}

		area += (x_right - x_left) * y_length;
	}

	printf("Test case #%d\nTotal explored area: %.2lf\n\n", caze, area);
}

int main() {
	int n;
	int caze = 0;
	while (true) {
		scanf("%d", &n);
		if (n == 0) break;
		caze++;
		for (int i = 0; i < n; i++) {
			scanf("%lf %lf %lf %lf", &x1[i], &y1[i], &x2[i], &y2[i]);
		}
		processCase(caze, n);
	}
}