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POJ 2002 Squares (哈希)
题意:给定 n 个点,问有能组成多少个正方形。
析:通过直接桥梁两个顶点,然后再算另外两个,再通过哈希进行查找另外两个,这里我先是用的map,竟然卡过了3400ms多,后来改成哗哈希,900ms,哈希我也是用STL中的容器来写的,list,先枚举的那两个点是相邻的,然后再通过旋转90度,去计算另外两个。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const int mod = 99991; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } list<P> Hash[mod]; P a[maxn]; P solve(const P &lhs, const P &rhs){ int detx = rhs.first - lhs.first; int dety = rhs.second - lhs.second; return P(lhs.first-dety, lhs.second+detx); } int calc(const P &p){ return (p.first * p.first + p.second * p.second) % mod; } int main(){ while(scanf("%d", &n) == 1 && n){ for(int i = 0; i < mod; ++i) Hash[i].clear(); for(int i = 0; i < n; ++i){ scanf("%d %d", &a[i].first, &a[i].second); Hash[calc(a[i])].push_back(a[i]); } int ans = 0; for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ if(i == j) continue; P p3 = solve(a[i], a[j]); int x = calc(p3); bool ok = false; for(list<P> :: iterator it = Hash[x].begin(); it != Hash[x].end(); ++it) if(*it == p3){ ok = true; break; } if(!ok) continue; P p4 = solve(p3, a[i]); x = calc(p4); ok = false; for(list<P> :: iterator it = Hash[x].begin(); it != Hash[x].end(); ++it) if(*it == p4){ ok = true; break; } if(!ok) continue; ++ans; } } printf("%d\n", ans / 4); } return 0; }
POJ 2002 Squares (哈希)
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