poj 2002 Squares
题目连接
http://poj.org/problem?id=2002
Squares
Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
题意:给出一些点的集合求能组成正方形的个数。
思路:直接枚举的话$n = 1000\ \ \ O(n^4)$ 会超时。考虑换个思路:已知两个点,由正方形的几何特性
求出另外两个点的坐标,判断是否在点的集合里(哈希,二分,set。。。)都可我用的哈希。
具体先对所有的点按横坐标,纵坐标从小到大排序。统计个数$tot$,最后的答案即为${tot/2}$(正方形的对称性)
那么问题来了,如何已知两个点,求另外两个点呢?
现在给出公式:记$A(x_1,y_1)\ \ \ B(x_2,y_2) \ \ \overrightarrow {AB} =(x_2-x_1,y_2-y_1)$
另外两个点$C(x_3, y_3)\ \ \ D(x_4,y_4)\ \ \ $
$C:\ \ \ x_3 = y_1 - y_2 + x_1\ \ y_3 = x_2 - x_1 + y_1 $
$D:\ \ \ x_4= y_1 - y_2 + x_2\ \ y_4 = x_2 - x_1 + y_2 $
证明其实很简单记$\overrightarrow{X} = (a,b)$逆时针旋转$\beta$度得到$\overrightarrow{Y}(x,y)$
有:$x = acos\beta-bsin\beta\ \ y = asin\beta+bcos\beta$ (由三角函数的几何意义易得)
那么$\overrightarrow {AC} = (x_3-x_1,y_3-y_1)$
$x_3-x_1=(x_2-x_1)cos\beta - (y_2-y_1)sin\beta$
$y_3-y_1=(x_2-x_1)sin\beta + (y_2-y_1)cos\beta$ 其中$\beta = 90^0$
所以$x_3 = y_1-y_2+x_1\ \ y_3=x_2-x_1+y_1$
$\overrightarrow {BD}$同理(注意向量的方向和旋转方向)
原谅我孱弱的语文水平写的太挫了凑合看吧/(ㄒoㄒ)/~~
#include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<vector> #include<map> using std::map; using std::abs; using std::sort; using std::pair; using std::vector; using std::multimap; #define pb(e) push_back(e) #define sz(c) (int)(c).size() #define mp(a, b) make_pair(a, b) #define all(c) (c).begin(), (c).end() #define iter(c) __typeof((c).begin()) #define cls(arr, val) memset(arr, val, sizeof(arr)) #define cpresent(c, e) (find(all(c), (e)) != (c).end()) #define rep(i, n) for(int i = 0; i < (int)n; i++) #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i) const int N = 40007; const int INF = 0x3f3f3f3f; struct P { int x, y; P() {} P(int i , int j) :x(i), y(j) {} inline bool operator<(const P &k) const { return x == k.x ? y < k.y : x < k.x; } }A[N]; struct Hash_Set { int tot, head[N]; struct edge { int x, y, next; }G[N]; inline void init() { tot = 0, cls(head, -1); } inline void insert(P &k) { int u = abs(k.x + k.y) % N; G[tot] = (edge){ k.x, k.y, head[u] }; head[u] = tot++; } inline bool find(P k) { int u = abs(k.x + k.y) % N; for(int i = head[u]; ~i; i = G[i].next) { edge &e = G[i]; if(k.x == e.x && k.y == e.y) return true; } return false; } }hash; int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); freopen("out.txt", "w+", stdout); #endif int n, x1, x2, x3, x4, y1, y2, y3, y4, ans; while(~scanf("%d", &n), n) { hash.init(); rep(i, n) { scanf("%d %d", &A[i].x, &A[i].y); hash.insert(A[i]); } sort(A, A + n); ans = 0; for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { x1 = A[i].x, y1 = A[i].y; x2 = A[j].x, y2 = A[j].y; x3 = y1 - y2 + x1, y3 = x2 - x1 + y1; x4 = y1 - y2 + x2, y4 = x2 - x1 + y2; if(hash.find(P(x3, y3)) && hash.find(P(x4, y4))) ans++; } } printf("%d\n", ans >> 1); } return 0; }