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HDU 6055 Regular polygon (暴力)
题意,二维平面上给N个整数点,问能构成多少个不同的正多边形。
析:容易得知只有正四边形可以使得所有的顶点为整数点。所以只要枚举两个点,然后去查找另外两个点就好。
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 = 500 + 10; const LL mod = 1e9 + 7; 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; } map<P, int> mp; P p[maxn]; P solve(const P &p1, const P &p2){ int detx = p2.first - p1.first; int dety = p2.second - p1.second; return P(p1.first-dety, p1.second+detx); } int main(){ while(scanf("%d", &n) == 1){ mp.clear(); for(int i = 0; i < n; ++i){ scanf("%d %d", &p[i].first, &p[i].second); mp[p[i]] = true; } int ans = 0; for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ if(i == j) continue; P p1 = solve(p[i], p[j]); if(!mp.count(p1)) continue; P p2 = solve(p1, p[i]); if(!mp.count(p2)) continue; ++ans; } } printf("%d\n", ans/4); } return 0; }
HDU 6055 Regular polygon (暴力)
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