hdu 5080 2014ACM/ICPC鞍山K题 polya计数

标签：blog   io   os   ar   for   sp   on   2014   art   

首先,中心点是可以直接算出来的

把所有的坐标相加再除n就可以

然后枚举一个不靠近中心的点，枚举它绕中心点旋转的角度，只要枚举50次就可以了

计算出当前枚举的的角度是否能形成一个置换群

计算循环节，再用polya定理算个数

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)			memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("out.txt","r",stdin)
#define write           freopen("out2.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-6;
const double pi = acos(-1.0);
const int maxn = 55;
const int mod = 1000000007;

int n, m, c;
int nx[maxn];
int ny[maxn];
double cita[maxn];
double r[maxn];
double cx, cy;
int a[maxn];
int b[maxn];
int edge[maxn][maxn];

i64 gcd(i64 _a, i64 _b)
{
	if (!_a || !_b)
	{
		return max(_a, _b);
	}
	i64 _t;
	while ((_t = _a % _b))
	{
		_a = _b;
		_b = _t;
	}
	return _b;
}

i64 ext_gcd(i64 _a, i64 _b, i64 &_x, i64 &_y)
{
	if (!_b)
	{
		_x = 1;
		_y = 0;
		return _a;
	}
	i64 _d = ext_gcd(_b, _a % _b, _x, _y);
	i64 _t = _x;
	_x = _y;
	_y = _t - _a / _b * _y;
	return _d;
}

i64 invmod(i64 _a, i64 _p)
{
	i64 _ans, _y;
	ext_gcd(_a, _p, _ans, _y);
	_ans < 0 ? _ans += _p : 0;
	return _ans;
}

double gao(double x,double y){
	if (abs(x) < eps){
		if (y>0){
			return pi / 2.0;
		}
		else{
			return pi + pi / 2.0;
		}
	}
	else if (x >= 0 && y >= 0){
		return atan(y / x);
	}
	else if (x <= 0 && y >= 0){
		x = -x;
		return pi - atan(y / x);
	}	
	else if (x <= 0 && y <= 0){
		x = -x;
		y = -y;
		return pi + atan(y / x);
	}
	else {
		y = -y;
		return 2 * pi - atan(y / x);
	}
}

int find(double tcita,double tr){
	if (tcita > 2 * pi){
		tcita -= 2 * pi;
	}
	double tx = cx + tr*cos(tcita);
	double ty = cy + tr*sin(tcita);
	for (int i = 1; i <= n; i++){
		if (abs(tx - nx[i]) < eps && abs(ty-ny[i]) < eps){
			return i;
		}
	}
	return -1;
}

bool isint(double temp){
	int t2 = temp;
	temp -= t2;
	if (temp < eps) {
		return true;
	}
	else{
		return false;
	}
}

bool can(){
	int now, to;
	for (int x = 1; x <= n; x++){
		for (int y = 1; y <= n; y++){
			now = b[x];
			to = b[y];
			if (edge[x][y] != edge[now][to]){
				return false;
			}
		}
	}
	return true;
}

bool vis[maxn];

int count(){
	MM(vis, 0);
	int re = 0;
	int now, to;
	for (int x = 1; x <= n; x++){
		if (!vis[x]){
			now = x;
			vis[now] = true;
			re++;
			while (true){
				to = b[now];
				if (vis[to]){
					break;
				}
				else{
					vis[to] = true;
					now = to;
				}
			}
		}
	}
	return re;
}

i64 pow_mod(int x,int temp){
	i64 re = 1;
	for (int i = 1; i <= temp; i++){
		re *= x;
		re %= mod;	
	}
	return re;
}

i64 start(){
	cx = 0.0;
	cy = 0.0;
	for (int i = 1; i <= n; i++){
		cx += nx[i];
		cy += ny[i];
	}
	cx /= n;
	cy /= n;
	double tx, ty;
	for (int i = 1; i <= n; i++){
		tx = nx[i] - cx;
		ty = ny[i] - cy;
		r[i] = sqrt(tx*tx + ty*ty);
		cita[i] = gao(tx, ty);
	}
	double spin;
	i64 ans = 0;
	i64 sg =0;
	int id;
	for (int i = 1; i <= n; i++){
		if (abs(cx - nx[i]) > eps || abs(cy - ny[i]) > eps){
			id = i;
			break;
		}
	}
	for (int i = 1; i <= n; i++){
		spin = cita[i] - cita[id];
		if (abs(r[i] - r[id]) < eps){
			bool no = false;
			for (int x = 1; x <= n; x++){
				a[x] = find(cita[x] + spin, r[x]);
				if (a[x] == -1){
					no = true;
					break;
				}
				b[a[x]] = x;
			}
			if (no) continue;
			if (can()){
				sg++;
				ans += pow_mod(c, count());
				ans %= mod;
			}
		}
	}
	ans *= invmod(sg, mod);
	ans %= mod;
	return ans;
}

int main(){
	int T;
	cin >> T;
	while (T--){
		cin >> n >> m >> c;
		for (int i = 1; i <= n; i++){
			cin >> nx[i] >> ny[i];
		}
		for (int i = 0; i <= n; i++){
			for (int j = 0; j <= n; j++){
				edge[i][j] = 0;
			}
		}
		int now, to;
		for (int i = 1; i <= m; i++){
			cin >> now >> to;
			edge[now][to] = edge[to][now] = 1;
		}
		if (n == 1){
			cout << c << endl;
			continue;
		}
		i64 ans = start();	
		cout << ans << endl;
	}
	return 0;
}

hdu 5080 2014ACM/ICPC鞍山K题 polya计数

标签：blog   io   os   ar   for   sp   on   2014   art   

原文地址：http://blog.csdn.net/zz_1215/article/details/40429513