POJ 2074 Line of Sight（求直线交点 + 思维）

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\n", ans);
    }

    return 0;
}