usaco 5.2 Electric Fences(아날로그 후퇴)

Electric Fences Kolstad & Schrijvers
Farmer John has decided to construct electric fences. He has fenced his fields into a number of bizarre shapes and now must find the optimal place to locate the electrical supply to each of the fences.
A single wire must run from some point on each and every fence to the source of electricity. Wires can run through other fences or across other wires. Wires can run at any angle. Wires can run from any point on a fence (i.e., the ends or anywhere in between) to the electrical supply.
Given the locations of all F (1 <= F <= 150) fences (fences are always parallel to a grid axis and run from one integer gridpoint to another, 0 <= X,Y <= 100), your program must calculate both the total length of wire required to connect every fence to the central source of electricity and also the optimal location for the electrical source.
The optimal location for the electrical source might be anywhere in Farmer John's field, not necessarily on a grid point.

PROGRAM NAME: fence3

INPUT FORMAT

The first line contains F, the number of fences. F subsequent lines each contain two X,Y pairs each of which denotes the endpoints of a fence.

SAMPLE INPUT (file fence3.in)

3
0 0 0 1
2 0 2 1
0 3 2 3

OUTPUT FORMAT

On a single line, print three space-separated floating point numbers, each with a single decimal place. Presume that your computer's output library will round the number correctly.
The three numbers are:

the X value of the optimal location for the electricity,

the Y value for the optimal location for the electricity, and

the total (minimum) length of the wire required.

SAMPLE OUTPUT (file fence3.out)

1.0 1.6 3.7

제목: n선 구간을 드릴게요. 이 점에서 n선 구간까지의 거리와 최단을 구하세요.
분석: 이 문제를 보자마자 나는 아날로그 퇴화로 할 생각을 했다. 선을 점으로 삼으면 마점을 구하는 문제가 되었다. 즉, 점에서 선까지의 거리의 함수를 하나만 쓰면 되고 나머지는 마점을 구하는 틀로 하면 된다.
그러나 이 문제는 선이 하나의 점인 상황을 주의하여 나로 하여금 한 번 괴롭혔다
코드:

/*
ID: 15114582
PROG: fence3
LANG: C++
*/
#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
const int mm=222;
const double eps=1e-2;
typedef double diy;
struct point
{
    diy x,y;
    point(){}
    point(diy _x,diy _y):x(_x),y(_y){}
}s[mm],t[mm];
point Vector(point s,point t)
{
    return point(t.x-s.x,t.y-s.y);
}
diy CrossProduct(point P,point Q)
{
    return P.x*Q.y-P.y*Q.x;
}
diy DotProduct(point P,point Q)
{
    return P.x*Q.x+P.y*Q.y;
}
diy MultiCross(point P,point Q,point R)
{
    return CrossProduct(Vector(Q,P),Vector(Q,R));
}
diy MultiDot(point P,point Q,point R)
{
    return DotProduct(Vector(Q,P),Vector(Q,R));
}
diy Dis(point P,point Q)
{
    return sqrt((P.x-Q.x)*(P.x-Q.x)+(P.y-Q.y)*(P.y-Q.y));
}
diy PointSegDis(point P,point Q,point R)
{
    if(Dis(P,Q)<eps)return Dis(P,R);
    if(MultiDot(Q,P,R)<-eps)return Dis(P,R);
    if(MultiDot(P,Q,R)<-eps)return Dis(Q,R);
    return fabs(MultiCross(P,Q,R)/Dis(P,Q));
}
diy AllDis(point o,int n)
{
    diy ret=0;
    while(n--)ret+=PointSegDis(s[n],t[n],o);
    return ret;
}
int dx[]={0,0,-1,1};
int dy[]={-1,1,0,0};
int i,n;
int main()
{
    freopen("fence3.in","r",stdin);
    freopen("fence3.out","w",stdout);
    point o,tmp;
    diy T,delta,ans,ret;
    bool flag;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;++i)
            scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&t[i].x,&t[i].y);
        o=s[0];
        ans=AllDis(o,n);
        flag=1;
        for(T=100,delta=0.99;T>eps;T*=delta,flag=1)
            while(flag)
                for(flag=i=0;i<4;++i)
                {
                    tmp.x=o.x+dx[i]*T;
                    tmp.y=o.y+dy[i]*T;
                    ret=AllDis(tmp,n);
                    if(ans>ret)
                    {
                        ans=ret;
                        o=tmp;
                        flag=1;
                    }
                }
        printf("%.1lf %.1lf %.1lf
",o.x,o.y,ans);
    }
    return 0;
}