hdu 5080 2014ACM/ICPC鞍山K题 polya计数
首先,中心点是能够直接算出来的
把全部的坐标相加再除n就能够
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。仅仅要枚举50次就能够了
计算出当前枚举的的角度能否形成一个置换群
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;
}
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