hunnu11325（点到多边形的距离）

首先判断点是否在多边形上，然后求点到多边形每条边的最短距离

代码如下：

#include<iostream>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<>
#include<>
#include<>
#include<>
#include<>


#define N 105
#define inf 0xffffffff
#define eps 1e-9
#define pi acos(-1.0)
#define P system("pause")
using namespace std;
struct point 
{
     double x,y;     
     point(double a,double b){x=a;y=b;}
     point (){};
}p[N],A;
int n;
point operator - (point A,point B)
{
           return point(,);       
}
double dot(point A,point B)
{
    return *+*;       
}
double cross(point A,point B)
{
    return **;       
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    return 1;    
}
int isonsegment(point a1,point a2)
{
     if((==&&==)||(==&&==)) return 1;    
     return dcmp(cross(a1-A,a2-A))==0&&dcmp(dot(a1-A,a2-A))<0;
}
int isonpoly()
{
     
     int i,wn=0;
     for(i=0;i<n;i++)
     {           
          if(isonsegment(p[i],p[(i+n)%n]))
             return 1;
          int k=dcmp(cross(p[(i+1)%n]-p[i],A-p[i]));
          int d1=dcmp(p[i].);
          int d2=dcmp(p[(i+1)%n].);
          if(k>0&&d1<=0&&d2>0) wn++;
          if(k<0&&d2<=0&&d1>0)  wn--;                            
     }    
     if(wn!=0) return 1;
     return 0;
}
double length(point B)
{
    return sqrt(dot(B,B));       
}
double dist(point a1,point a2 )
{
    if(==&&==) return length(A-a1 );       
    point v1=a2-a1,v2=A-a1,v3=A-a2;
    if(dcmp(dot(v1,v2)<0)) return length(v2);
    else if(dcmp(dot(v1,v3))>0) return length(v3);
    else return fabs(cross(v1,v2))/length(v1);
}

int main()
{
//freopen("","r",stdin);
//freopen("","w",stdout);
    int m;
  
    while(scanf("%d",&n)&&n)
    {
          int i;  
          scanf("%lf%lf",&,&);                  
          for(i=0;i<n;i++)
             scanf("%lf%lf",&p[i].x,&p[i].y);
          if(isonpoly()){printf("0.00\n");continue;}   //判断点是否在多边形上
          double ret=inf;
          for(i=0;i<n;i++)
                ret =min(ret, dist(p[i],p[(i+1)%n]));
          printf("%.2lf\n",ret);
          }                     
//    P;                               
    return 0;    
}