題目鏈接: http://acm.hdu.edu.cn/showproblem.php?pid=6055

題意: 給出 n 組坐標 x, y, 輸出其中的正多邊形個數 . 其中 x, y 均為整數.

思路: x, y 為整數, 所以只存在正方形, 不會有其他正多邊形 . 那麽只需要枚舉正方形的對角線即可 .

代碼:

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6
 
 using namespace std;
 7 
 8 const int MAXN = 1e3 + 10;
 9 const int H = 1e4 + 7;
10 
11 struct node{
12     int x, y, next;
13 }gel[MAXN];
14 
15 ll ans;
16 int head[H];
17 int indx, n;
18 int ptx[MAXN], pty[MAXN];
19 
20 void init(void){
21     memset(head, -1, sizeof(head));
22     indx = 0;
23     ans = 0
 
;
24 }
25 
26 void add(int x, int y){
27     int h = (x * x + y * y) % H;
28     gel[indx].x = x;
29     gel[indx].y = y;
30     gel[indx].next = head[h];
31     head[h] = indx++;
32 }
33 
34 bool find(int x, int y){
35     int h = (x * x + y * y) % H;
36     for(int i = head[h]; i != -1; i = gel[i].next){
 
37         if(x == gel[i].x && y == gel[i].y) return true;
38     }
39     return false;
40 }
41 
42 int main(void){
43     while(~scanf("%d", &n)){
44         init();
45         for(int i = 0; i < n ; i++){
46             scanf("%d%d", &ptx[i], &pty[i]);
47             add(ptx[i], pty[i]);
48         }
49         for(int i = 0; i < n; i++){
50             for(int j = 0; j < n; j++){
51                 if(i == j) continue;
52                 int x1 = ptx[i] - (pty[i] - pty[j]);
53                 int y1 = pty[i] + (ptx[i] - ptx[j]);
54                 int x2 = ptx[j] - (pty[i] - pty[j]);
55                 int y2 = pty[j] + (ptx[i] - ptx[j]);
56                 if(find(x1, y1) && find(x2, y2)) ans++;
57             }
58         }
59         printf("%lld\n", ans >> 2);
60     }
61     return 0;
62 }

hdu6055(求正方形個數)