Form a Square
题意就是 判断 给你四个点,能否组成一个正方形
要点:
格式很重要, 很重要!!!
数据很小,直接暴力
四个点判断是否为正方形,只需将所有可能的边长度算出来,然后选其中最短的边作为正方形的边长,进行比较,看有没有符合的四条边
1 #include<iostream> 2 #include<algorithm> 3 #include<cmath> 4 using namespace std; 5 const int N = 4; 6 typedef struct aa{ 7 int x, y; 8 }AA; 9 AA a[N]; 10 11 double dis(AA b, AA c) 12 { 13 double len = sqrt((b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y)); 14 return len; 15 } 16 int main() 17 { 18 int t, num, cnt = 1; 19 cin >> t; 20 int T = t; 21 while(T--) 22 { 23 for(int i = 0; i < 4; i++) 24 { 25 cin >> a[i].x >> a[i].y; 26 } 27 double ll = 2010; 28 num = 0; 29 for(int i = 0; i < 4; i++) 30 { 31 for(int j = i+1; j < 4; j++) 32 { 33 double dist = dis(a[i], a[j]); 34 if(dist == ll) 35 { 36 num++; 37 } 38 else if(dist < ll) 39 { 40 ll = dist; 41 num = 1; 42 } 43 } 44 } 45 if(num == 4) 46 { 47 if(cnt != t) 48 cout << "Case " << "" << cnt++ << ":" << endl << "Yes" << endl << endl; 49 else 50 cout << "Case " << "" << cnt++ << ":" << endl << "Yes"; 51 } 52 else 53 { 54 if(cnt != t) 55 cout << "Case " << "" << cnt++ << ":" << endl << "No" << endl << endl; 56 else 57 cout << "Case " << "" << cnt++ << ":" << endl << "No"; 58 } 59 60 } 61 return 0; 62 }