HDU 1542 Atlantis（矩形面积并）

最新推荐文章于 2024-08-17 03:33:08 发布

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最新推荐文章于 2024-08-17 03:33:08 发布

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原文链接：http://www.cnblogs.com/jzssuanfa/p/7082012.html

本文介绍了解决 HDU1542 Atlantis 题目的方法，采用扫描线算法计算多个矩形的并集面积。通过使用线段树或直接暴力扫描来处理矩形集合，并详细展示了实现过程。

HDU 1542 Atlantis

题目链接

题意：给定一些矩形，求面积并

思路：利用扫描线，因为这题矩形个数不多，直接暴力扫就能够了。假设数据大。就要用线段树

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 205;
const int M = 100005;
const double eps = 1e-8;

int n, vis[N], hn;
double hash[N];

struct Line {
	double l, r, y;
	int flag;
	Line() {}
	Line(double l, double r, double y, int flag) {
		this->l = l;
		this->r = r;
		this->y = y;
		this->flag = flag;
	}
} line[N];

bool cmp(Line a, Line b) {
	return a.y < b.y;
}

int get(double x) {
	return lower_bound(hash, hash + hn, x) - hash;
}

int main() {
	int cas = 0;
	while (~scanf("%d", &n) && n) {
		double x1, x2, y1, y2; hn = 0;
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < n; i++) {
			scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
			line[i * 2] = Line(x1, x2, y1, 1);
			line[i * 2 + 1] = Line(x1, x2, y2, -1);
			hash[hn++] = x1; hash[hn++] = x2;
		}
		n *= 2;
		sort(line, line + n, cmp);
		sort(hash, hash + hn);
		hn = 1;
		for (int i = 1; i < n; i++) {
			if (fabs(hash[i] - hash[i - 1]) < eps) continue;
			hash[hn++] = hash[i];
		}
		double ans = 0;
		for (int i = 0; i < n; i++) {
			int l = get(line[i].l), r = get(line[i].r);
			double len = 0;
			for (int j = 0; j < hn - 1; j++) if (vis[j] > 0) len += (hash[j + 1] - hash[j]);
			if (i) ans += len * (line[i].y - line[i - 1].y);
			for (int j = l; j < r; j++) vis[j] += line[i].flag;
		}
		printf("Test case #%d\n", ++cas);
		printf("Total explored area: %.2lf\n\n", ans);
	}
	return 0;
}

转载于:https://www.cnblogs.com/jzssuanfa/p/7082012.html