bzoj 1488: [HNOI2009]图的同构 （置换+dfs）

Description

Input

Output

Sample Input

Sample Output

HINT

Source

题解：置换＋dfs

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 103
#define p 997
#define LL long long 
using namespace std;
int n,num[N],k[N];
LL cnt,c[N][N],ans,d[N],jc[N],inv[N],a[N],g[N][N],b[N*N];
LL quickpow(LL num,int x)
{
    LL ans=1; LL base=num%p;
    while (x) {
        if (x&1) ans=ans*base%p;
        x>>=1;
        base=base*base%p;
    }
    return ans;
}
void init(int n)
{
    for (int i=0;i<=n;i++) c[i][0]=1;
    for (int i=1;i<=n;i++)
     for (int j=1;j<=i;j++) c[i][j]=(c[i-1][j]+c[i-1][j-1])%p;
    jc[0]=1;
    for (int i=1;i<=n;i++) jc[i]=jc[i-1]*i%p;
    for (int i=1;i<=n;i++) inv[i]=quickpow(jc[i],p-2);
    b[0]=1;
    for (int i=1;i<=n*n;i++) b[i]=b[i-1]*2%p;
}
int gcd(int x,int y){
    if (g[x][y]) return g[x][y];
    int r;
    while (y){
        r=x%y;
        x=y; y=r;
    }
    return g[x][y]=x;
}
void dfs(int x,int now)
{
    if (!now) {
        int n1=x-1; int sum=0;
        for (int i=1;i<=n1;i++)
         for (int j=i+1;j<=n1;j++) sum+=gcd(a[i],a[j]);
        for (int i=1;i<=n1;i++) sum+=a[i]/2;
        LL tot=1; int cnt=n;
        for (int i=1;i<=n1;i++)
         tot=tot*c[cnt][a[i]]%p,cnt-=a[i];
        for (int i=1;i<=n;i++)
         if (k[i]) tot=tot*inv[k[i]]%p;
        for (int i=1;i<=n1;i++) tot=tot*jc[a[i]-1]%p;
        ans=(ans+tot*b[sum])%p;
        //cout<<tot<<" "<<sum<<endl;
        return;
    }
    for (int i=a[x-1];i<=now;i++) {
        k[i]++; a[x]=i;
        dfs(x+1,now-i);
        k[i]--;
    }
}
int main()
{
    scanf("%d",&n);
    if (n==1) {
        printf("1\n");
        return 0; 
    }
    a[0]=1;
    init(n); dfs(1,n);
    ans=ans*quickpow(jc[n],p-2)%p;
    printf("%lld\n",ans);
}