HDOJ 5365 Run 【几何】

HDOJ 5365 Run 【正多边形】

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define clr(c) memset(c, 0, sizeof(c));
#define debug(x) cout<<"debug "<<x<<endl;
#define pi acos(-1.0)
const int INF = 0x3f3f3f3f;
typedef long long ll;
const int MAXL = 25;
int n;

typedef struct point{
int x, y;
bool operator < (const point& p) const{
if(x == p.x) return y < p.y;
else return x < p.x;
}
bool operator > (const point& p) const{
return p < *this;
}
}p;

p p1, p2, p3, p4;
p point[MAXL];
double l1, l2, l3, l4, l5, l6;
int cnt;

inline double length(p p1, p p2){
return (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y);
}

void Resolve(){
cnt = 0;
sort(point, point+n);
for(int i = 0; i < n-3; i++)
for(int j = i+1; j < n-2; j++)
for(int k = j+1; k < n-1; k++)
for(int l = k+1; l < n; l++){
p1 = point[i];
p2 = point[j];
p3 = point[k];
p4 = point[l];
l1 = length(p1, p2);
l2 = length(p1, p3);
l3 = length(p2, p4);
l4 = length(p3, p4);
l5 = length(p1, p4);
l6 = length(p2, p3);
if(l1 == l2 && l2 == l3 && l3 == l4 && l5 == l6) cnt++;
}
printf("%d\n", cnt);
}

int main(){
while(~scanf("%d", &n)){
for(int i = 0; i < n; i++){
scanf("%d%d", &point[i].x, &point[i].y);
}
Resolve();
}
return 0;
}