UVALive - 7251 2015 ICPC Hefei

题目链接：UVALive - 7251

题意：

统计带标号 n 个点可以组成多少种不同的有环连通图，答案
对质数 $(2^{15}\times 4641+1)$ 取模，其中每条边可以染成 $m$ 种颜
色

sol:

考虑算出联通图的方案数，再减去树的方案个数。

1.$n$个点树的方案个数为$n^{n-2}$
2.考虑任意图的指数生成函数
$G(x)={\sum }_{n\ge 0}\frac{(m+1{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}}{n!}{x}^n$，即对所有的边考虑染色.
3. 设连通图的指数生成函数为$F(x)$
$G(x)={\sum }_{k\ge 1}\frac{F_k(x)}{k!}=\exp (F(x))$
4.对$G(x)$求对数即可，详情见$2015$金策论文

code:

#include

using namespace std;
typedef long long ll;

const int maxn = 4e5+10;
//不声明为常量慢一倍
const int P = 152076289;
const int mod = 152076289;
const ll g = 127;

void Add(ll& x,ll y){
    x += y;
    if(x>=mod) x-= mod;
}

void Mul(ll& x,ll y){
    x *= y;
    if(x>=mod) x%=mod;
}

ll qpow(ll a,ll b){
    ll ret = 1;
    while(b){
        if(b&1) Mul(ret,a);
        b >>= 1;
        Mul(a,a);
    }
    return ret;
}

inline ll Inv(ll a){
    return qpow(a,mod-2);
}

ll inv[maxn],fac[maxn],finv[maxn];

void init(){
    inv[0] = inv[1] = fac[0] = fac[1] = 1;
    for(int i = 2;i<maxn;i++) inv[i] = inv[P % i] * (P - P/i) % P;
    for(int i = 2;i<maxn;i++) fac[i] = fac[i - 1] * i % mod;
    finv[0] = finv[1] = 1;
    for(int i = 2;i<maxn;i++) finv[i] = finv[i-1] * inv[i] % mod;
}

struct NTT{
    int rev[maxn],N,L;

    void init_rev(int n){
        for(N = 1,L=0;N<=n;N<<=1,L++);
        for(int i=1;i<N;i++) rev[i] = (rev[i >> 1] >> 1 | ((i&1) << (L-1)));
    }

    void DFT(ll a[],int flag){
        for(int i = 0;i<N;i++)
            if(i<rev[i]) swap(a[i],a[rev[i]]);
        for(int l = 2;l<=N;l <<=1 ){
            ll wn;
            if(flag == 1) wn = qpow(g,(P-1)/l);
            else wn = qpow(g,P-1-(P-1)/l);
            for(int k = 0;k<N;k += l){
                ll w = 1;
                ll x,y;
                for(int j = k;j < k + l/2 ;j++){
                    x = a[j];
                    y = a[j + l/2] * w % P;
                    a[j] = (x + y) % P;
                    a[j + l/2] = (x - y +P) % P;
                    Mul(w,wn);
                }
            }
        }
        if(flag == -1){
            ll inv = Inv(N);
            for(int i = 0; i<N ;i++) a[i] = a[i] * inv % P;
        }
    }
}ntt;

ll A[maxn],B[maxn],C[maxn],D[maxn];
ll f[maxn],G[maxn];

void poly_inv(int deg,ll* a,ll * b){
    if(deg == 1){
        b[0] = Inv(a[0]);
        return ;
    }
    poly_inv((deg+1)>>1,a,b); ntt.init_rev(deg<<1);
    for(int i = 0;i<deg;i++) C[i] = a[i];
    for(int i = deg;i<ntt.N;i++) C[i] = b[i] = 0;
    ntt.DFT(C,1); ntt.DFT(b,1);
    for(int i = 0;i<ntt.N;i++) 
        b[i] = ((2LL - C[i] * b[i] % P) + P) % P * b[i] % P;
    ntt.DFT(b,-1);
    for(int i = deg;i<ntt.N;i++) b[i] = 0;
}

void Direv(int n,ll* A,ll* B,int flag){
    if(flag == 1){
        for(int i = 1;i<n;i++) B[i-1] = A[i] * i % mod;
        B[n - 1] = 0;
    }
    else{
        for(int i = 1;i<n;i++) B[i] = A[i-1] * inv[i] % mod;
        B[0] = 0;
    }
}

void get_log(int n,ll* a,ll* b){
    Direv(n,a,A,1);
    poly_inv(n,a,B);
    ntt.DFT(A,1); ntt.DFT(B,1);
    for(int i = 0;i<ntt.N;i++) A[i] = A[i] * B[i] % mod;
    ntt.DFT(A,-1);
    Direv(n,A,b,-1);
    for(int i = 0;i<ntt.N;i++) A[i] = B[i] = 0;
    for(int i = n;i<ntt.N;i++) b[i] = 0;
}

int main(){
    init();
    int T;
    scanf("%d",&T);
    for(int cas = 1;cas<=T;cas++){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 0;i <= n + 100;i++){
            ll c2 = (ll)i * (i-1) / 2;
            f[i] = qpow(m+1,c2);
            // f[i] = f[i] * Inv(fac[i]) % mod;
            // if(i<10) cout<
            f[i] = f[i] * finv[i] % mod;
        }
        // puts("");
        get_log(n+10,f,G);
        ll ans = G[n] * fac[n] % mod;
        ll ret = 0;
        if(n >= 2) ret = qpow(n,n-2) * qpow(m,n-1) % mod;
        ans = (ans + mod - ret) % mod;
        printf("Case #%d: %lld\n",cas,ans);
    }
    return 0;
}