HDU 1542 Atlantis（矩形面积并）

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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，由于这题矩形个数不多，直接暴力扫就可以了，如果数据大，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;}

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