BZOJ 1004 Cards(Burnside引理+DP)

因为有着色数的限制，故使用Burnside引理。

添加一个元置换(1,2,,,n)形成m+1种置换，对于每个置换求出循环节的个数，

每个循环节的长度。

则ans=sigma(f(i))/(m+1) %p  (1<=i<=m+1).

其中f(i)是第i种置换下的不动点个数。

可以用dp来求出f(i), 设第i个置换的循环节个数为T, 令dp[i][j][k]表示前i个循环节中使用了j个红色，k个蓝色的不动点个数。进行一次n^3的DP即可。

最后m+1模p意义下的逆元不再叙述。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF (LL)<<
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

struct Per{int a[];}per[];
int dp[][][], n, vis[], num[];

int get_loop(int x)
{
    int cnt=;
    mem(vis,); mem(num,);
    FOR(i,,n) {
        if (vis[i]) continue;
        ++cnt;
        int now=i;
        while (vis[now]==) vis[now]=, now=per[x].a[now], ++num[cnt];
    }
    return cnt;
}
int extend_gcd(int a, int b, int &x, int &y)
{
    if (a==&&b==) return -;
    if (b==){x=; y=; return a;}
    int d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
int mod_reverse(int a, int n)
{
    int x, y, d=extend_gcd(a,n,x,y);
    if (d==) return (x%n+n)%n;
    else return -;
}
int main ()
{
    int sr, sb, sg, m, p;
    LL ans=;
    scanf("%d%d%d%d%d",&sr,&sb,&sg,&m,&p);
    n=sr+sb+sg;
    FOR(i,,m) FOR(j,,n) scanf("%d",&per[i].a[j]);
    FOR(j,,n) per[m+].a[j]=j;
    FOR(i,,m+) {
        int t=get_loop(i);
        mem(dp,);
        dp[][][]=;
        int sum=;
        for (int j=; j<=t; ++j) FOR(k,,sr) FOR(l,,sb) {
            sum+=num[j];
            if (k+l>sum) continue;
            if (sum-k-l>=num[j]) dp[j][k][l]=dp[j-][k][l];
            if (k>=num[j]) dp[j][k][l]=(dp[j][k][l]+dp[j-][k-num[j]][l])%p;
            if (l>=num[j]) dp[j][k][l]=(dp[j][k][l]+dp[j-][k][l-num[j]])%p;
        }
        ans=(ans+dp[t][sr][sb])%p;
    }
    ans=ans*mod_reverse(m+,p)%p;
    printf("%lld\n",ans);
    return ;
}