HDU 1542 Atlantis
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; }
时间: 2024-10-28 23:49:58