Fermat Point in Quadrangle（hdu 3694 求两直线交点

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3694

画几个图应该就可以知道凸四边形就是对角线交点 凹四边形就是凹进去的那个点

so 只要枚举四个点以及对角线交点 找个minn就可以

求两直线交点模板：

double cross(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}
bool getcross(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4,double &tx,double &ty)
{
    double A1=y2-y1;
    double B1=x1-x2;
    double C1=cross(x1,y1,x2,y2);
    double A2=y4-y3;
    double B2=x3-x4;
    double C2=cross(x3,y3,x4,y4);
    if(fabs(cross(A1,B1,A2,B2))<1e-6)return 0;
    tx=cross(C1,B1,C2,B2)/cross(A1,B1,A2,B2);
    ty=cross(A1,C1,A2,C2)/cross(A1,B1,A2,B2);
    return 1;
}
int main()
{
    int i,j;
    double x1,x2,x3,x4,y1,y2,y3,y4;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))
    {
        double ty,tx;
        if(getcross(x1,y1,x2,y2,x3,y3,x4,y4,tx,ty))cout<<tx<<" "<<ty<<endl;
    }
}

代码：

#include<bits/stdc++.h>
using namespace std;
double dis(double x2,double y2,double x1,double y1)
{
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}
double minn,x[7],y[7];
void judge(double x1,double y1)
{
    int i;
    double sum=0;
    for(i=1;i<=4;i++)
    {
        sum+=dis(x1,y1,x[i],y[i]);
    }
    if(sum<minn)minn=sum;
}
double cross(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}
bool getcross(int a,int b,int c,int d,double &tx,double &ty)//求两直线交点
{
    double A1=y[b]-y[a];
    double B1=x[a]-x[b];
    double C1=cross(x[a],y[a],x[b],y[b]);
    double A2=y[d]-y[c];
    double B2=x[c]-x[d];
    double C2=cross(x[c],y[c],x[d],y[d]);
    if(fabs(cross(A1,B1,A2,B2))<1e-6)return 0;
    tx=cross(C1,B1,C2,B2)/cross(A1,B1,A2,B2);
    ty=cross(A1,C1,A2,C2)/cross(A1,B1,A2,B2);
    return 1;
}
int main()
{
    int i,j;
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x[1],&y[1],&x[2],&y[2],&x[3],&y[3],&x[4],&y[4]),x[1]>=0)
    {
        minn=0x3f3f3f3f;
        double ty,tx;
        for(i=1;i<=4;i++)judge(x[i],y[i]);
        if(getcross(1,2,3,4,tx,ty))judge(tx,ty);
        if(getcross(1,3,2,4,tx,ty))judge(tx,ty);
        if(getcross(1,4,2,3,tx,ty))judge(tx,ty);
        printf("%.4f\n",minn);
    }
}

原文地址：https://www.cnblogs.com/ydw--/p/11369439.html