【SRM 565 UnknownTree】计数 分类讨论

一个有N + 3个点的树，告诉你123号点到其他点的距离，求合法的边权为正整数的树个数。

#include 
#include 
#include 
#include 
#include 
#include 
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
typedef double DB;
const int mod = 1000000009, N = 60;
LL ans; int n, a[N], b[N], c[N], par[N], p0[N];
class UnknownTree {
public:
	void Calc(int *a, int *b, int *c, int Dab, int Dac) {
        if (!Dab || !Dac) return ;
        a[n] = b[n + 1] = c[n + 2] = 0;
		a[n + 1] = b[n] = Dab, a[n + 2] = c[n] = Dac, b[n + 2] = c[n + 1] = Dab + Dac;
		LL sum = 1;
		memset(par, 0, sizeof(par));
		Rep(j, 0, n - 1) {
			Rep(k, 0, n + 2)
				if (a[j] > a[k] && a[j] - a[k] == b[j] - b[k] && a[j] - a[k] == c[j] - c[k]) par[j] ++;
			if (par[j]) { (sum *= par[j]) %= mod; continue ; }
			Rep(k, 0, n + 2) if (j != k && a[j] == a[k] && b[j] == b[k] && c[j] == c[k]) sum = 0;
			if (a[j] + b[j] == Dab && c[j] - a[j] == Dac) continue ;
			if (a[j] + c[j] == Dac && b[j] - a[j] == Dab) continue ;
			sum = 0;
		}
		(ans += sum) %= mod;
	}
	void Work(int *a, int *b, int *c) {
		if (a[0] > b[0]) {
			if (a[0] < c[0]) Calc(a, b, c, a[0] - b[0], c[0] - a[0]);
		} else {
			if (a[0] > c[0]) Calc(a, b, c, b[0] - a[0], a[0] - c[0]);
			else Calc(a, b, c, b[0] - a[0], c[0] - a[0]);
			Calc(a, b, c, b[0] - a[0], a[0] + c[0]);
		}
		if (c[0] > a[0]) Calc(a, b, c, a[0] + b[0], c[0] - a[0]);
	}
	int getCount(vector  A, vector  B, vector  C) {
		n = A.size();
		Rep(i, 0, n - 1) a[i] = A[i], b[i] = B[i], c[i] = C[i];
		Rep(i, 0, n - 1) {
            int Dab, Dac, Dbc;
			a[n + 1] = b[n] = Dab = a[i] + b[i];
			a[n + 2] = c[n] = Dac = a[i] + c[i];
			b[n + 2] = c[n + 1] = Dbc = b[i] + c[i];
			memset(par, 0, sizeof(par));
            LL sum = 1; bool fl = 0;
			Rep(j, 0, n - 1) if (j != i) {
				Rep(k, 0, n + 2)
					if (a[j] > a[k] && a[j] - a[k] == b[j] - b[k] && a[j] - a[k] == c[j] - c[k]) par[j] ++;
				if (par[j]) { (sum *= par[j]) %= mod; continue ; }
				Rep(k, 0, n + 2) if (j != k && a[j] == a[k] && b[j] == b[k] && c[j] == c[k]) fl = 1;
				if (a[j] < a[i] && (a[j] + b[j] != Dab || a[j] + c[j] != Dac)) fl = 1;
				if (b[j] < b[i] && (a[j] + b[j] != Dab || c[j] + b[j] != Dbc)) fl = 1;
				if (c[j] < c[i] && (a[j] + c[j] != Dac || c[j] + b[j] != Dbc)) fl = 1;
				if (a[j] >= a[i] && b[j] >= b[i] && c[j] >= c[i]) fl = 1;
			}
			if (!fl) (ans += sum) %= mod;
		}
        Rep(j, 0, n - 1) {
			Rep(k, 0, n - 1)
				if (a[j] > a[k] && a[j] - a[k] == b[j] - b[k] && a[j] - a[k] == c[j] - c[k]) p0[j] ++;
            if (!p0[j]) { swap(a[j], a[0]), swap(b[j], b[0]), swap(c[j], c[0]); break ; }
        }
		Work(a, b, c);
		Work(b, a, c);
		Work(c, a, b);
		return (int)ans;
	}
};