poj 1410 Intersection （判矩形和线段相交。。细节多）

【题目大意】：给出一条线段和一个矩形，判断线段和矩形是否相交。

 

【解题思路】：判给出的线段与矩形的每条线段是否相交（包括相交和非规范相交）。

1、内含也算相交（WA一次）       2、看了discuss，发现题目给出矩形的点并不一定是左上角的点和右下角的点（WA2次）

 

【代码】：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;#define eps 1e-8struct Point{    double x, y;    Point() {}    Point(double a, double b)    {        x = a, y = b;    }}point[20];struct Line{    Point a, b;    Line() {}    Line(Point x, Point y)    {        a = x, b = y;    }}line,line1[20];inline int sig(double k){    return k < -eps ? -1 : k > eps;}inline double det(double x1, double y1, double x2, double y2){    return x1 * y2 - x2 * y1;}inline double xmult(Point o, Point a, Point b){    return det(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);}inline double dotdet(double x1, double y1, double x2, double y2){    return x1 * x2 + y1 * y2;}inline double dot(Point o, Point a, Point b){    return dotdet(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);}inline int between(Point o, Point a, Point b){    return sig(dot(o, a, b));}inline int intersect1(Point a, Point b, Point c, Point d) {    double s1, s2, s3, s4;    int d1 = sig(s1 = xmult(a, b, c));    int d2 = sig(s2 = xmult(a, b, d));    int d3 = sig(s3 = xmult(c, d, a));    int d4 = sig(s4 = xmult(c, d, b));    if ((d1^d2) == -2 && (d3^d4) == -2)    {        return 1;    }    if (d1 == 0 && between(c, a, b) <= 0) return 2;    if (d2 == 0 && between(d, a, b) <= 0) return 2;    if (d3 == 0 && between(a, c, d) <= 0) return 2;    if (d4 == 0 && between(b, c, d) <= 0) return 2;    return 0;}inline int intersect(Line u, Line v){    return intersect1(u.a, u.b, v.a, v.b);}int main(){    int T;    double p,q,m,n,p1,q1,m1,n1;    scanf("%d",&T);    while (T--)    {        scanf("%lf%lf",&p1,&q1);        scanf("%lf%lf",&m1,&n1);        point[1]=Point(m1,n1);        point[0]=Point(p1,q1);        line=Line(point[0],point[1]);        scanf("%lf%lf%lf%lf",&p,&q,&m,&n);        if (p>m && q<n)        {            swap(p,m);            swap(q,n);        }        else        if (p>m && q>n) swap(m,p);        else        if (p<m && q<n) swap(q,n);        if ((p1>=p && p1<=m && q1<=q && q1>=n && m1>=p && m1<=m && n1<=q && n1>=n))        {            printf("T\n");            continue;        }        point[2]=Point(p,q);        point[3]=Point(m,q);        point[4]=Point(p,n);        point[5]=Point(m,n);        line1[0]=Line(point[2],point[3]);        line1[1]=Line(point[2],point[4]);        line1[2]=Line(point[3],point[5]);        line1[3]=Line(point[4],point[5]);        int k1,k2,k3,k4;        k1=intersect(line,line1[0]);        k2=intersect(line,line1[1]);        k3=intersect(line,line1[2]);        k4=intersect(line,line1[3]);        //cout << k1 << " " << k2 << " " << k3 << " " << k4 << endl;        if (k1==0 && k2==0 && k3==0 && k4==0)        {            printf("F\n");        }        else printf("T\n");    }    return 0;}

 

【运行结果】：