题目:传送门
题意:有一个房子(用一条线段表示),从马路(用一条线段表示)看房子,有许多障碍物(用线段表示),问在马路上连续的一段线段都能看到完整的房子最长是多长。
思路:就是求线段交点,然后,把那些交点按 x 排序,最后求答案即可。 有个陷阱就是障碍物可能不在房子和马路之间。
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <set> #include <string> #include <math.h> #define LL long long #define mem(i, j) memset(i, j, sizeof(i)) #define rep(i, j, k) for(int i = j; i <= k; i++) #define dep(i, j, k) for(int i = k; i >= j; i--) #define pb push_back #define make make_pair #define INF INT_MAX #define inf LLONG_MAX #define PI acos(-1) using namespace std; struct Point{ double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } }; const int N = 110; const double eps = 1e-8; int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } Point operator + (Point A, Point B) { return Point(A.x + B.x, A.y + B.y); } Point operator - (Point A, Point B) { return Point(A.x - B.x, A.y - B.y); } Point operator * (Point A, double p) { return Point(A.x * p, A.y * p); } Point operator / (Point A, double p) { return Point(A.x / p, A.y / p); } bool operator == (Point A, Point B) { return dcmp(A.x-B.x) == 0 && dcmp(A.y - B.y) == 0; } bool operator < (Point A, Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y); } double Cross(Point A, Point B) { /// 叉积 return A.x * B.y - A.y * B.x; } double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; /// 点积 } double Length(Point A) { return sqrt(Dot(A, A)); } double Polar_Angle(Point A) { return atan2(A.y, A.x); } /// 向量极角 double Angle(Point A, Point B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量转角,逆时针 Point rorate(Point A, double rad) { ///向量旋转,逆时针 return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); } Point nomal(Point A) { /// 向量的单位法向量 double len = Length(A); return Point(-A.y / len, A.x / len); } bool Onseg(Point p, Point a1, Point a2) { /// 判断点 p 是否在线段 a1a2 上(含端点) return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) <= 0; } bool SPI(Point a1, Point a2, Point b1, Point b2) { /// 判断线段a1a2与线段b1b2是否相交 return max(a1.x,a2.x) >= min(b1.x,b2.x) && max(b1.x,b2.x) >= min(a1.x,a2.x) && max(a1.y,a2.y) >= min(b1.y,b2.y) && max(b1.y,b2.y) >= min(a1.y,a2.y) && dcmp(Cross(b1 - a2, a1 - a2)) * dcmp(Cross(b2 - a2, a1 - a2)) <= 0 && dcmp(Cross(a1 - b2, b1 - b2)) * dcmp(Cross(a2 - b2, b1 - b2)) <= 0; } double DistanceToSegment(Point p, Point A, Point B) { /// 求点 p 到线段 AB 的最短距离 if(A == B) return Length(p - A); Point v1 = B - A, v2 = p - A, v3 = p - B; if(dcmp(Dot(v1, v2)) < 0) return Length(v2); else if(dcmp(Dot(v1, v3)) > 0) return Length(v3); else return fabs(Cross(v1, v2)) / Length(v1); } inline Point GetLineIntersection(const Point P, const Point v, const Point Q, const Point w) {///求直线p + v*t 和 Q + w*t 的交点,需确保有交点,v和w是方向向量 Point u = P - Q; double t = Cross(w, u) / Cross(v, w); return P + v * t; } Point P[N]; int main() { double x1, x2, y; while(scanf("%lf %lf %lf", &x1, &x2, &y)) { if(dcmp(x1) == 0 && dcmp(x2) == 0 && dcmp(y) == 0) break; Point sta1 = Point(x1, y), sta2 = Point(x2, y); scanf("%lf %lf %lf", &x1, &x2, &y); Point eda1 = Point(x1, y), eda2 = Point(x2, y); int n; scanf("%d", &n); vector < Point > Q; rep(i, 0, n - 1) { scanf("%lf %lf %lf", &x1, &x2, &y); if(dcmp(y - sta1.y) >= 0 || dcmp(y - eda1.y) <= 0) continue; Point p1 = Point(x1, y), p2 = Point(x2, y); x1 = GetLineIntersection(sta2, sta2 - p1, eda1, eda1 - eda2).x; x2 = GetLineIntersection(sta1, sta1 - p2, eda1, eda1 - eda2).x; if(dcmp(x1 - x2) >= 0) continue; Q.pb(Point(x1, x2)); } Q.pb(Point(eda2.x, eda2.x)); sort(Q.begin(), Q.end()); double L = eda1.x, ans = 0; for(int i = 0; i < Q.size(); i++) { Point v = Q[i]; x1 = v.x; x2 = v.y; if(x1 - L > ans) ans = x1 - L; if(dcmp(x2 - L) > 0) { L = x2; if(dcmp(L - eda2.x) > 0) break; } } if(dcmp(ans) == 0) puts("No View"); else printf("%.2f ", ans); } return 0; }