CSU 1805 Three Capitals(BEST定理)

题意：给你三个点A,B,G,AB之间有a条双向边，AG之间有b条边，BG之间c有条边

问你从A出发然后回到A的方法个数，要求走过所有的边，即求欧拉回路的个数

 

思路：从A出发回到A，如果是有向边，那么即求有向图的欧拉回路个数，可用best定理求，所以这里考虑将无向图转化为有向图，有向图的欧拉回路中每个顶点的入度一定等于出度，可以在AB之间枚举A的入边，那么其他的所有情况都能求出来了，假设AB之间有x条A->B的边，AG之间有y条A->G的边，BG之间有z条B->G的边，那么同一形态的有向树的总数有C(a,x)*C(b,y)*C(c,z)，其中C是组合数

 

best定理如下:以s为顶点计数有向图的欧拉回路

#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<iostream>
#include<algorithm>
typedef long long ll;
const ll mod = 1e9 + 7;
const ll maxn = 1e5 + 10;
using namespace std;

ll A[5][5];
ll fac[maxn];
ll deg[5];
ll a, b, c;

ll qmod(ll x, ll n, ll m) {
    ll ans = 1;
    while(n) {
        if(n & 1) ans = (ans * x) % m;
        x = x * x % m;
        n >>= 1;
    }
    return ans;
}

ll C(ll n, ll m) { return fac[n] * (qmod(fac[m], mod - 2, mod) * qmod(fac[n - m], mod - 2, mod) % mod) % mod; }

int main() {
    fac[0] = 1;
    for(ll i = 1; i < maxn; i++) fac[i] = (fac[i - 1] * i) % mod;
    while(scanf("%lld %lld %lld", &a, &b, &c) != EOF) {
        ll ans = 0;
        if(((a + b) & 1) || ((c + b) & 1) || ((a + c) & 1)) { printf("0\n"); continue; }
        for(ll x = 0; x <= a; x++) {
            ll atob = x, btoa = a - x;
            ll atoy = (a + b) / 2 - x, ytoa = b + atob - (a + b) / 2;
            ll ytob = (b + c) / 2 - ytoa;
            ll btoy = c - ytob;
            if(atoy < 0 || ytoa < 0 || btoy < 0 || ytob < 0) continue;
            deg[0] = btoa + ytoa;
            deg[1] = atob + ytob;
            deg[2] = atoy + btoy;
            for(ll i = 0; i < 3; i++) A[i][i] = deg[i];
            A[1][2] = -btoy; A[2][1] = -ytob;
            ll res = (C(a, x) * C(b, atoy) % mod) * C(c, ytob) % mod;
            ll det = ((A[1][1] * A[2][2] % mod) - (A[1][2] * A[2][1] % mod) + mod) % mod;
            res = res * det % mod;
            res = res * deg[0] % mod;
            for(ll i = 0; i < 3; i++) res = res * fac[deg[i] - 1] % mod;
            ans = (ans + res) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}