[HDOJ5365]Run

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5365

众所周知，整点是围不成正三角形正五边形正六边形的，所以我们只需暴力出它可以围成几个正四边形。

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <cmath>
 7 #include <cctype>
 8 #include <queue>
 9 #include <map>
10 #include <set>
11 #include <stack>
12 #include <list>
13 #include <vector>
14
15 using namespace std;
16
17 typedef struct Node {
18     int x;
19     int y;
20 };
21 Node po[22];
22 int edge[11];
23 int n;
24
25 int dis(int x, int y) {
26     return x * x + y * y;
27 }
28 int solve() {
29     memset(edge, 0, sizeof(edge));
30     int ans = 0;
31     for(int i = 0; i < n; i++) {
32         for(int j = i+1; j < n; j++) {
33             for(int p = j+1; p < n; p++) {
34                 for(int q = p+1; q < n; q++) {
35                     edge[0] = dis(po[i].x-po[j].x, po[i].y-po[j].y);
36                     edge[1] = dis(po[i].x-po[p].x, po[i].y-po[p].y);
37                     edge[2] = dis(po[i].x-po[q].x, po[i].y-po[q].y);
38                     edge[3] = dis(po[j].x-po[p].x, po[j].y-po[p].y);
39                     edge[4] = dis(po[j].x-po[q].x, po[j].y-po[q].y);
40                     edge[5] = dis(po[p].x-po[q].x, po[p].y-po[q].y);
41                     sort(edge, edge+6);
42                     if(edge[0] == edge[1] && edge[1] == edge[2] && edge[2] == edge[3] && edge[4] == edge[5]) {
43                         ans++;
44                     }
45                 }
46             }
47         }
48     }
49     return ans;
50 }
51 int main() {
52     // freopen("in", "r", stdin);
53     while(~scanf("%d", &n)) {
54         for(int i = 0; i < n; i++) {
55             scanf("%d %d", &po[i].x, &po[i].y);
56         }
57         printf("%d\n", solve());
58     }
59 }