题意：从一条马路(线段)看对面的房子(线段)，问连续的能看到房子全部的最长区间
分析：自己的思路WA了：先对障碍物根据坐标排序，然后在相邻的障碍物的间隔找到区间，这样还要判断是否被其他障碍物遮挡住(哇
网上有很好的思路，先对每条线段找到阴影的端点，然后根据坐标排序，求和左端点的距离的最大值，这样省去线段相交的判断。
trick点应该就是障碍物的位置随意，可能在房子和马路的外面。
/************************************************
* Author :Running_Time
* Created Time :2015/11/2 星期一 20:33:38
* File Name :POJ_2074.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x) { //三态函数，减少精度问题
if (fabs (x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
struct Point { //点的定义
double x, y;
Point () {}
Point (double x, double y) : x (x), y (y) {}
Point operator + (const Point &r) const { //向量加法
return Point (x + r.x, y + r.y);
}
Point operator - (const Point &r) const { //向量减法
return Point (x - r.x, y - r.y);
}
Point operator * (double p) const { //向量乘以标量
return Point (x * p, y * p);
}
Point operator / (double p) const { //向量除以标量
return Point (x / p, y / p);
}
bool operator < (const Point &r) const { //点的坐标排序
return x < r.x || (x == r.x && y < r.y);
}
bool operator == (const Point &r) const { //判断同一个点
return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
}
};
typedef Point Vector; //向量的定义
Point read_point(void) { //点的读入
double x, y;
scanf ("%lf%lf", &x, &y);
return Point (x, y);
}
double dot(Vector A, Vector B) { //向量点积
return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B) { //向量叉积
return A.x * B.y - A.y * B.x;
}
double polar_angle(Vector A) { //向量极角
return atan2 (A.y, A.x);
}
double length(Vector A) { //向量长度，点积
return sqrt (dot (A, A));
}
double angle(Vector A, Vector B) { //向量转角，逆时针，点积
return acos (dot (A, B) / length (A) / length (B));
}
Vector rotate(Vector A, double rad) { //向量旋转，逆时针
return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A) { //向量的单位法向量
double len = length (A);
return Vector (-A.y / len, A.x / len);
}
Point line_line_inter(Point p, Vector V, Point q, Vector W) { //两直线交点，参数方程
Vector U = p - q;
double t = cross (W, U) / cross (V, W);
return p + V * t;
}
double point_to_line(Point p, Point a, Point b) { //点到直线的距离，两点式
Vector V1 = b - a, V2 = p - a;
return fabs (cross (V1, V2)) / length (V1);
}
double point_to_seg(Point p, Point a, Point b) { //点到线段的距离，两点式
if (a == b) return length (p - a);
Vector 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);
}
Point point_line_proj(Point p, Point a, Point b) { //点在直线上的投影，两点式
Vector V = b - a;
return a + V * (dot (V, p - a) / dot (V, V));
}
bool can_seg_seg_inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交，两点式
double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool can_line_seg_inter(Point a1, Point a2, Point b1, Point b2) { //判断直线与线段相交，两点式
double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1);
return dcmp (c1 * c2) <= 0;
}
bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上，两点式
return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
} struct Line2 {
double x1, x2;
Line2 () {}
Line2 (double x1, double x2) : x1 (x1), x2 (x2) {}
bool operator < (const Line2 &r) const {
return x1 < r.x1;
}
}; struct Line {
Point a, b;
Line () {}
Line (Point a, Point b) : a (a), b (b) {}
}; int main(void) {
Line h, p;
double x1, x2, y;
while (scanf ("%lf%lf%lf", &x1, &x2, &y) == 3) {
if (dcmp (x1) == 0 && dcmp (x2) == 0 && dcmp (y) == 0) break;
h.a = Point (x1, y); h.b = Point (x2, y);
scanf ("%lf%lf%lf", &x1, &x2, &y);
p.a = Point (x1, y); p.b = Point (x2, y);
int n; scanf ("%d", &n);
vector<Line2> xs;
for (int i=0; i<n; ++i) {
scanf ("%lf%lf%lf", &x1, &x2, &y);
if (dcmp (y - h.a.y) >= 0 || dcmp (y - p.a.y) <= 0) continue;
Point p1 = Point (x1, y), p2 = Point (x2, y);
x1 = line_line_inter (h.b, h.b - p1, p.b, p.b - p.a).x;
x2 = line_line_inter (h.a, h.a - p2, p.b, p.b - p.a).x;
if (dcmp (x1 - x2) >= 0) continue;
xs.push_back (Line2 (x1, x2));
}
xs.push_back (Line2 (p.b.x, p.b.x));
sort (xs.begin (), xs.end ());
double left = p.a.x, ans = 0;
for (int i=0; i<xs.size (); ++i) {
x1 = xs[i].x1; x2 = xs[i].x2;
if (x1 - left > ans) ans = x1 - left;
if (dcmp (x2 - left) > 0) {
left = x2;
if (dcmp (left - p.b.x) > 0) break;
}
}
if (dcmp (ans) == 0) puts ("No View");
else printf ("%.2f\n", ans);
} //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0;
}