POJ 2002 Squares （二分查找，简单几何）

quares

题意比较简单，就是在一堆点用找能组成正方形的点，看能找出多少组。

开始没思路，怕超时。上网找了找，基本思路是枚举两个点，查找另外两个看存不存在。这个就是简单的几何，因为点是排过序的，我们只需要找到两个点在左面（如果y值一样就在下面）的情况，但这样还会多算一遍，因为一个正方形要有两条边在左面（或一条左一条下），所以要除以2。

查找有用二分查找的，有用hash的。因为stl里有二分查找函数，所以好写些。

 

#include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <string> #include <cstdio> #include <climits> #include <queue> #include <map> #include <list> #include <vector> using namespace std; struct node { int x, y; } p[1001]; bool op(const struct node &xx, const struct node &yy) { if (xx.x == yy.x) return xx.y < yy.y; else return xx.x < yy.x; } int main() { int points; while (scanf("%d", &points), points) { int sum = 0; for (int i = 0; i < points; ++i) scanf("%d%d", &p[i].x, &p[i].y); sort(p, p + points, op); for (int i = 0; i < points; ++i) { for (int j = i + 1; j < points; ++j) { node p0, p1; p0.x = p[i].x + p[j].y - p[i].y; p0.y = p[i].y + p[i].x - p[j].x; p1.x = p[j].x + p[j].y - p[i].y; p1.y = p[j].y + p[i].x - p[j].x; if (!binary_search(p, p+points, p0, op)) continue; if (!binary_search(p, p+points, p1, op)) continue; sum++; } } printf("%d/n", sum / 2); } return 0; }

时间1.5s多

 

刚才突然想到用set写，结果用时2.6s多……

#include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <string> #include <cstdio> #include <climits> #include <queue> #include <map> #include <set> #include <vector> using namespace std; struct node { int x, y; friend bool operator<(const node &a, const node &b) { if (a.x == b.x) return a.y < b.y; else return a.x < b.x; } }; int main() { int points; while (scanf("%d", &points), points) { int sum = 0; set<node> s; for (int i = 0; i < points; ++i) { node tmp; scanf("%d%d", &tmp.x, &tmp.y); s.insert(tmp); } set<node>::iterator ita, itb; for (ita = s.begin(); ita != s.end(); ita++) { for (itb = ita; itb != s.end(); itb++) { if (ita == itb) continue; node p0, p1; p0.x = ita->x + itb->y - ita->y; p0.y = ita->y + ita->x - itb->x; p1.x = itb->x + itb->y - ita->y; p1.y = itb->y + ita->x - itb->x; if (s.find(p0)== s.end()) continue; if (s.find(p1)==s.end()) continue; sum++; } } printf("%d/n", sum / 2); } return 0; }

 

后来发现可以一个正方形只算一次，时间降到了700ms+

 

#include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <string> #include <cstdio> #include <climits> #include <queue> #include <map> #include <list> #include <vector> using namespace std; struct node { int x, y; } p[1001]; bool op(const struct node &xx, const struct node &yy) { if (xx.x == yy.x) return xx.y < yy.y; else return xx.x < yy.x; } int main() { int points; while (scanf("%d", &points), points) { int sum = 0; for (int i = 0; i < points; ++i) scanf("%d%d", &p[i].x, &p[i].y); sort(p, p + points, op); for (int i = 0; i < points; ++i) { for (int j = i + 1; j < points; ++j) { if (p[i].x <= p[j].x && p[i].y >= p[j].y) { node p0, p1; p0.x = p[i].x + p[j].y - p[i].y; p0.y = p[i].y + p[i].x - p[j].x; p1.x = p[j].x + p[j].y - p[i].y; p1.y = p[j].y + p[i].x - p[j].x; if (!binary_search(p, p+points, p0, op)) continue; if (!binary_search(p, p+points, p1, op)) continue; sum++; } } } printf("%d/n", sum); } return 0; }