POJ 1269 Intersecting Lines(判断直线相交）

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题目地址：POJ 1269

直接套模板就可以了。。。实在不想自己写模板了。。。写的又臭又长。。。。不过这题需要注意的是要先判断是否有直线垂直X轴的情况。

代码如下：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define eqs 1e-10struct node{    double x, y;}point;int dcmp(double x, double y){    if(fabs(x-y)<=eqs)        return 1;    return 0;}node intersection(node u1, node u2, node v1, node v2){    node ret=u1;    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));    ret.x+=(u2.x-u1.x)*t;    ret.y+=(u2.y-u1.y)*t;    return ret;}int main(){    int n, i;    printf("INTERSECTING LINES OUTPUT\n");    scanf("%d",&n);    node f1, f2, f3, f4;    while(n--)    {        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&f1.x,&f1.y,&f2.x,&f2.y,&f3.x,&f3.y,&f4.x,&f4.y);        node a, b;        a.x=f2.x-f1.x;        a.y=f2.y-f1.y;        b.x=f4.x-f3.x;        b.y=f4.y-f3.y;        if(dcmp(f1.x,f2.x)&&dcmp(f3.x,f4.x))        {            if(f1.x==f3.x)            {                printf("LINE\n");            }            else            {                printf("NONE\n");            }            continue ;        }        else if(dcmp(f1.x,f2.x))        {            double k=b.y/b.x;            double b=f3.y-(f3.x*k);            double y=k*f1.x+b;            printf("POINT %.2lf %.2lf\n",f1.x,y);            continue ;        }        else if(dcmp(f3.x,f4.x))        {            double k=b.y/b.x;            double b=f1.y-(f1.x*k);            double y=k*f3.x+b;            printf("POINT %.2lf %.2lf\n",f3.x,y);            continue ;        }        double k1=a.y/a.x;        double k2=b.y/b.x;        double b1=f1.y-(k1*f1.x);        double b2=f3.y-(k2*f3.x);        if(dcmp(k1,k2))        {            if(dcmp(b1,b2))            {                printf("LINE\n");            }            else                printf("NONE\n");        }        else        {            node ff=intersection(f1,f2,f3,f4);            printf("POINT %.2lf %.2lf\n",ff.x,ff.y);        }    }    printf("END OF OUTPUT\n");    return 0;}

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