題意:給一個n和n個整數座標問這些點能組成幾個不同正三角形,正方形,正五邊形,正六邊形。
分析:由於座標都是整數,使用只可能有正方形,其他都不可能,那麼只要對於每四個不同的點,判斷2組對邊相等,兩條對角線也相等,臨邊也相等即可。
代碼:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int maxn = 25;
int d[2][maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++) scanf("%d%d",&d[0][i],&d[1][i]);
int ans=0;
for(int k1=1;k1<=n-3;k1++)
for(int k2=k1+1;k2<=n-2;k2++)
for(int k3=k2+1;k3<=n-1;k3++)
for(int k4=k3+1;k4<=n;k4++){
double b1=sqrt((d[0][k1]-d[0][k2])*(d[0][k1]-d[0][k2])*1.0+(d[1][k1]-d[1][k2])*(d[1][k1]-d[1][k2]));//這三組邊是對邊,對邊,對角線三種任意的順序
double b2=sqrt((d[0][k3]-d[0][k4])*(d[0][k3]-d[0][k4])*1.0+(d[1][k3]-d[1][k4])*(d[1][k3]-d[1][k4]));
double c1=sqrt((d[0][k1]-d[0][k3])*(d[0][k1]-d[0][k3])*1.0+(d[1][k1]-d[1][k3])*(d[1][k1]-d[1][k3]));
double c2=sqrt((d[0][k2]-d[0][k4])*(d[0][k2]-d[0][k4])*1.0+(d[1][k2]-d[1][k4])*(d[1][k2]-d[1][k4]));
double d1=sqrt((d[0][k1]-d[0][k4])*(d[0][k1]-d[0][k4])*1.0+(d[1][k1]-d[1][k4])*(d[1][k1]-d[1][k4]));
double d2=sqrt((d[0][k2]-d[0][k3])*(d[0][k2]-d[0][k3])*1.0+(d[1][k2]-d[1][k3])*(d[1][k2]-d[1][k3]));
if(b1==b2&&c1==c2&&d1==d2&&(b1==c1||b1==d1||c1==d1)) ans++; //分別判斷第一組對邊,第二組對邊,對角線,臨邊相等
}
printf("%d\n",ans);
}
}