luogu P1446 [HNOI2008]Cards burnside引理 置换 不动点

LINK：Cards

不太会burnside引理 而这道题则是一个应用。

首先 一个非常舒服的地方是这道题给出了m个本质不同的置换 然后带上单位置换就是m+1个置换.

burnside引理:

其中D(a_j)表示 在(a_j)这置换中的不动点的个数.

其实我们求出每个置换的不动点个数就行了.

循环很好求 每个循环都填一样的就是不动点了 直接dp一下即可.

code

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 1000000001
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d
",x)
#define putl(x) printf("%lld
",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-4
#define sq sqrt
#define S second
#define F first
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
	return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
	RE int x=0,f=1;RE char ch=gc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
	return x*f;

}
const int MAXN=66;
int n,m,mod;
int r,b,g,ans;
int vis[MAXN];
int c[MAXN];
int f[21][21][21],mark[MAXN];
inline int ksm(int b,int p)
{
	int cnt=1;
	while(p)
	{
		if(p&1)cnt=(ll)cnt*b%mod;
		b=(ll)b*b%mod;p=p>>1;
	}
	return cnt;
}
inline int calc()
{
	memset(f,0,sizeof(f));
	memset(mark,0,sizeof(mark));
	f[0][0][0]=1;int ww=0;
	rep(1,n,i)
	{
		if(!mark[i])
		{
			int cnt=1;
			mark[i]=1;
			int j=i;
			while(!mark[vis[j]])
			{
				j=vis[j];
				mark[j]=1;++cnt;
			}
			c[++ww]=cnt;
		}
	}
	rep(1,ww,T)
	{
		fep(r,0,i)fep(g,0,j)fep(b,0,k)
		{
			if(i>=c[T])f[i][j][k]=(f[i][j][k]+f[i-c[T]][j][k])%mod;
			if(j>=c[T])f[i][j][k]=(f[i][j][k]+f[i][j-c[T]][k])%mod;
			if(k>=c[T])f[i][j][k]=(f[i][j][k]+f[i][j][k-c[T]])%mod;
		}
	}
	return f[r][g][b];
}
int main()
{
	//freopen("1.in","r",stdin);
	get(r);get(b);get(g);
	get(m);n=r+b+g;get(mod);
	rep(1,m,i)
	{
		rep(1,n,j)get(vis[j]);
		ans=(ans+calc())%mod;
	}
	rep(1,n,j)vis[j]=j;
	ans=(ans+calc())%mod;
	ans=(ll)ans*ksm(m+1,mod-2)%mod;
	put(ans);return 0;
}