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  • POJ 3304 Segments (线段和直线相交 + 思维)

    题目: 传送门

    题意: 给你n条线段的两个端点,问所有线段投影到一条直线上,这些投影至少相交于一点,就输出Yes!,否则就是 No!

    题解: 

    #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;
    
    const int N = 310;
    
    struct Point {
        double x, y;
        Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数
    };
    
    typedef Point Vector;
    /// 向量+向量=向量, 点+向量=向量
    Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
    ///点-点=向量
    Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
    ///向量*数=向量
    Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
    ///向量/数=向量
    Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }
    
    const double eps = 1e-10;
    int dcmp(double x) {
        if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
    }
    
    bool operator < (const Point& a, const Point& b) {
        return a.x == b.x ? a.y < b.y : a.x < b.x;
    }
    
    bool operator == (const Point& a, const Point &b) {
        return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
    }
    
    double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } /// 点积
    double Length(Vector A) { return sqrt(Dot(A, A)); } /// 计算向量长度
    double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A、B夹角
    double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } /// 叉积
    Vector Rotate(Vector A, double rad) { /// 向量旋转, rad 是弧度, 逆时针旋转 rad
        return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
    }
    
    Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) { /// 求两直线交点,请确保 P + tv 和 Q + tw 有唯一交点。 当且仅当Cross(v, w)非0
        Vector u = P - Q;
        double t = Cross(w, u) / Cross(v, w);
        return P + v * t;
    }
    
    bool SegmentProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断线段是否相交
        double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);
        double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
        return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
    }
    
    bool OnSegment(Point p, Point a1, Point a2) { /// 点p是否在线段a1,a2上(不包含端点)
        return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
    }
    
    bool LSProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断直线b1b2是否和线段a1a2相交
        return dcmp(Cross(b1 - a1, b2 - a1)) * dcmp(Cross(b1 - a2, b2 - a2)) <= 0;
    }
    
    Point P[N], Q[N];
    int n;
    
    bool judge(Point a, Point b) {
        if(dcmp(Length(b - a)) == 0) return false;
         rep(i, 1, n) if(LSProperInsection(P[i], Q[i], a, b) == false) return false;
         return true;
    }
    
    void solve() {
        scanf("%d", &n);
        rep(i, 1, n) scanf("%lf %lf %lf %lf", &P[i].x, &P[i].y, &Q[i].x, &Q[i].y);
        rep(i, 1, n) rep(j, 1, n)
            if(judge(P[i], P[j]) || judge(P[i], Q[j]) || judge(Q[i], P[j]) || judge(Q[i], Q[j]))
                { puts("Yes!"); return ; }
        puts("No!"); return ;
    }
    
    int main() {
        int _; scanf("%d", &_);
        while(_--) solve();
        return 0;
    }
    一步一步,永不停息
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  • 原文地址:https://www.cnblogs.com/Willems/p/12337434.html
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