POJ 3384 Feng Shui（半平面交向内推进求最远点对）

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>

 using namespace std ;

 struct node
 {
     double x,y ;
 }p[110],temp[110],newp[110];
 int n,newn ;
 double a,b,c,r ;

 void getlinee(node x,node y)
 {
     a = y.y-x.y ;
     b = x.x-y.x ;
     c = y.x*x.y-x.x*y.y ;
 }
 node intersect(node x,node y)
 {
     double u = fabs(a*x.x+b*x.y+c) ;
     double v = fabs(a*y.x+b*y.y+c) ;
     node t ;
     t.x = (x.x*v+y.x*u)/(u+v) ;
     t.y = (x.y*v+y.y*u)/(u+v) ;
     return t ;
 }
 void cut()
 {
     int cutn = 0 ;
     for(int i = 1 ; i <= newn ; i++)
    {
        if(newp[i].x*a+newp[i].y*b+c >= 0)
            temp[++cutn] = newp[i] ;
        else
        {
            if(newp[i-1].x*a+newp[i-1].y*b+c > 0)
                temp[++cutn] = intersect(newp[i-1],newp[i]) ;
            if(newp[i+1].x*a+newp[i+1].y*b+c > 0)
                temp[++cutn] = intersect(newp[i+1],newp[i]) ;
        }
    }
    for(int i = 1 ; i <= cutn ; i++)
        newp[i] = temp[i] ;
    newp[cutn+1] = newp[1] ;
    newp[0] = newp[cutn] ;
    newn = cutn ;
    //printf("newn%d = %d\n",cnt++,newn) ;
 }
 void solve()
 {
     for(int i = 1 ; i <= n ; i ++)
    {
        newp[i] = p[i] ;
    }
    newp[n+1] = p[1] ;
    newp[0] = p[n] ;
    newn = n ;
    for(int i = 1 ; i <= n ; i++)
    {
        node t,t1,t2 ;
        t.x = p[i+1].y-p[i].y ;
        t.y = p[i].x-p[i+1].x ;
        double k = r/sqrt(t.x*t.x+t.y*t.y) ;
        t.x *= k ;
        t.y *= k ;
        t1.x = t.x+p[i].x ;
        t1.y = t.y+p[i].y ;
        t2.x = t.x+p[i+1].x ;
        t2.y = t.y+p[i+1].y ;
        getlinee(t1,t2) ;
        cut() ;
    }
 }
 int main()
 {
     while(~scanf("%d %lf",&n,&r))
    {
        for(int i = 1 ; i <= n ; i++)
            scanf("%lf %lf",&p[i].x,&p[i].y) ;
        p[n+1] = p[1] ;
        solve() ;
        int t1 = 0,t2 = 0 ;
        double maxx = 0.0 ;
        for(int i = 1 ; i <= newn ; i++)
        {
            for(int j = i+1 ; j <= newn ; j++)
            {
                double dis = sqrt((newp[i].x-newp[j].x)*(newp[i].x-newp[j].x)+(newp[i].y-newp[j].y)*(newp[i].y-newp[j].y)) ;
                if(dis > maxx)
                {
                    maxx = dis ;
                    t1 = i ;
                    t2 = j ;
                }
            }
        }
        printf("%.4lf %.4lf %.4lf %.4lf\n",newp[t1].x,newp[t1].y,newp[t2].x,newp[t2].y) ;
    }
     return 0 ;
 }