[JSOI2008] [洛谷P1227] 完美的对称

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
using namespace std;  
int x[20001],y[20001],n;  
double xx,yy;  
int read()  
{  
    char c=getchar();  
    int a=0;  
    bool flag=false;  
    while ((c<‘0‘||c>‘9‘)&&c!=‘-‘) c=getchar();  
    if (c==‘-‘)   
    {  
                flag=true;  
                c=getchar();  
    }  
    while (c>=‘0‘&&c<=‘9‘)   
    {  
          a=a*10+c-‘0‘;  
          c=getchar();  
    }  
    if (flag) return -a;else return a;  
}  
void swap(int i,int j)  
{  
     int t;  
     t=x[i]; x[i]=x[j]; x[j]=t;  
     t=y[i]; y[i]=y[j]; y[j]=t;  
}  
void quicksort(int head,int tail)  
{  
     int i=head,j=tail,mid=(i+j)/2;  
     while (i<j)  
     {  
           while (x[i]<x[mid]||x[i]==x[mid]&&y[i]<y[mid]) i++;  
           while (x[j]>x[mid]||x[j]==x[mid]&&y[j]>y[mid]) j--;  
           if (i<=j)  
           {  
                   swap(i,j);  
                   i++;  
                   j--;  
           }  
     }  
     if (head<j) quicksort(head,j);  
     if (i<tail) quicksort(i,tail);  
}  
int main()  
{  
    n=read();  
    for (int i=1;i<=n;i++)  
    {  
        x[i]=read();  
        y[i]=read();  
    }  
    quicksort(1,n);  
    xx=((double)(x[1]+x[n])/2);  
    yy=((double)(y[1]+y[n])/2);  
    for (int i=2;i<=n/2+1;i++)  
        if ((double)((x[i]+x[n-i+1])!=(xx*2))||(double)(y[i]+y[n-i+1])!=(yy*2))   
        {  
                                printf("This is a dangerous situation!");  
                                exit(0);  
        }  
    printf("V.I.P. should stay at (%.1lf,%.1lf).\n",xx,yy);  
    return 0;  
}