hdu 3694 Fermat Point in Quadrangle (数学)

题意：求一个点离给出的4个点最近，

思路：这个点是给出的点或者是对角线的交点。

求交点：a1x+b1y=c1 用两点式求出a1，b1，c1，然后用线性代数求解。克拉默法则都忘了，学妹实力一波教学。



#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
double x[10],y[10];
double ansx,ansy,min1;
double cross(double x1,double y1,double x2,double y2)
{
return x1*y2-x2*y1;
}
double sqr(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
bool getcross(int a,int b,int c,int d,double &dx,double &dy)
{
double a1=y[b]-y[a];/*a1x+b1y=c1 用两点式求出a1，b1，c1*/
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;
dx=cross(c1,b1,c2,b2)/cross(a1,b1,a2,b2);
dy=cross(a1,a2,c1,c2)/cross(a1,b1,a2,b2);
return 1;
}
void update(double x1,double y1)
{
double sum=0;
for(int i=1;i<=4;i++)sum+=sqr(x1,y1,x[i],y[i]);
if(sum<min1||min1<0)
{
min1=sum;
ansx=x1;
ansy=y1;
}
}
int main()
{
while(~scanf("%lf",&x[1]))
{
min1=-1;
if(x[1]<0)break;
scanf("%lf",&y[1]);
for(int i=2;i<=4;i++)scanf("%lf%lf",&x[i],&y[i]);
for(int i=1;i<=4;i++)update(x[i],y[i]);

double dx,dy;
if(getcross(1,2,3,4,dx,dy)){update(dx,dy);}/*线性代数求解*/
if(getcross(1,3,2,4,dx,dy))update(dx,dy);
if(getcross(1,4,2,3,dx,dy))update(dx,dy);
printf("%.4lf\n",min1);
}
return 0;
}