poj 1269 Intersecting Lines(计算几何)

Description

Input

Output

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

题意：有N组数据，每组数据分别给出两条直线的两个点，问这两条直线是否相交，不相交的话，如果共线则输出LINE，平行则输出NONE，相交的话，输出交点坐标

题解：三点共线的条件是：p1，p2，p3：(p2-p1)X(p3-p1)=0，没有交点就判断是否平行

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#define eps 1e-8

using namespace std;

struct point {
    double x,y;
} p[8];

///叉积
double ok(point a,point b,point c) {
    return (c.x-a.x)*(b.y-a.y)-(b.x-a.x)*(c.y-a.y);
}

//共线判断
bool Line(point a,point b,point c) {
    return ok(a,b,c)==0;
}

///交点判断
bool None() {
    return (p[1].y-p[2].y)*(p[3].x-p[4].x)==(p[1].x-p[2].x)*(p[3].y-p[4].y);
}

//求交点
point inter(point u1,point u2,point v1,point v2) {
    point 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() {
    //freopen("in.txt","r",stdin);
    int n;
    while(cin>>n) {
        printf("INTERSECTING LINES OUTPUT\n");
        while(n--) {
            for(int i=1; i<=4; i++)
                scanf("%lf%lff",&p[i].x,&p[i].y);
            if(Line(p[1],p[3],p[4])&&Line(p[2],p[3],p[4])) {
                printf("LINE\n");
                continue;
            }
            if(None()) {
                printf("NONE\n");
                continue;
            }
            point ans=inter(p[1],p[2],p[3],p[4]);
            printf("POINT %.2f %.2f\n",ans.x,ans.y);
        }
        printf("END OF OUTPUT\n");
    }
    return 0;
}

poj 1269 Intersecting Lines(计算几何)

原文：http://blog.csdn.net/acm_baihuzi/article/details/47299793