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-10
struct 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;
}