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开放寻址hash + 枚举n^2#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> using namespace std; const int N = 1000 + 5; const int MOD = 100007; struct Point { int x, y; Point(int _x = 0, int _y = 0) : x(_x), y(_y) {} } star[N], ht[MOD]; int hash(const Point &pt) { return (pt.x * pt.x + pt.y * pt.y) % MOD; } void insert(const Point &pt) { int idx = hash(pt); while (ht[idx].x != INT_MAX) idx = (idx + 1) % MOD; ht[idx] = pt; } bool search(const Point &pt) { int idx = hash(pt); while (ht[idx].x != INT_MAX) { if (ht[idx].x == pt.x && ht[idx].y == pt.y) return true; idx = (idx + 1) % MOD; } return false; } int main() { int n; while (scanf("%d", &n) == 1 && n) { for (int i = 0; i < MOD; i++) ht[i].x = INT_MAX; for (int i = 0; i < n; i++) scanf("%d%d", &(star[i].x), &(star[i].y)), insert(star[i]); int ans = 0; for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) { int x1 = star[i].x, x2 = star[j].x, y1 = star[i].y, y2 = star[j].y; Point pt3 = Point(x1 + y1 - y2, y1 + x2 - x1), pt4 = Point(x2 + y1 - y2, y2 + x2 - x1), pt5 = Point(x1 + y2 - y1, y1 + x1 - x2), pt6 = Point(x2 + y2 - y1, y2 + x1 - x2); ans += search(pt3) && search(pt4); ans += search(pt5) && search(pt6); } printf("%d\n", ans / 4); } return 0; } Followed by: Post your reply here: |
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