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-3
# define MOD 1000000007
# define INF (LL)1<<60
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=100005;
//Code begin...
 
struct Per{int a[65];}per[65];
int dp[65][65][65], n, vis[65], num[65];
 
int get_loop(int x)
{
    int cnt=0;
    mem(vis,0); mem(num,0);
    FOR(i,1,n) {
        if (vis[i]) continue;
        ++cnt;
        int now=i;
        while (vis[now]==0) vis[now]=1, now=per[x].a[now], ++num[cnt];
    }
    return cnt;
}
int extend_gcd(int a, int b, int &x, int &y)
{
    if (a==0&&b==0) return -1;
    if (b==0){x=1; y=0; 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==1) return (x%n+n)%n;
    else return -1;
}
int main ()
{
    int sr, sb, sg, m, p;
    LL ans=0;
    scanf("%d%d%d%d%d",&sr,&sb,&sg,&m,&p);
    n=sr+sb+sg;
    FOR(i,1,m) FOR(j,1,n) scanf("%d",&per[i].a[j]);
    FOR(j,1,n) per[m+1].a[j]=j;
    FOR(i,1,m+1) {
        int t=get_loop(i);
        mem(dp,0);
        dp[0][0][0]=1;
        int sum=0;
        for (int j=1; j<=t; ++j) FOR(k,0,sr) FOR(l,0,sb) {
            sum+=num[j];
            if (k+l>sum) continue;
            if (sum-k-l>=num[j]) dp[j][k][l]=dp[j-1][k][l];
            if (k>=num[j]) dp[j][k][l]=(dp[j][k][l]+dp[j-1][k-num[j]][l])%p;
            if (l>=num[j]) dp[j][k][l]=(dp[j][k][l]+dp[j-1][k][l-num[j]])%p;
        }
        ans=(ans+dp[t][sr][sb])%p;
    }
    ans=ans*mod_reverse(m+1,p)%p;
    printf("%lld
",ans);
    return 0;
}