本文主要是介绍POJ 1269 Intersecting Lines(判断直线相交），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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");}elseprintf("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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