CF1229D

题面

给定\(n\)个长度为\(k\)的置换，对于每个子段，求出通过使用零次或多次这些置换可以从初始排列\((1,2,\dots,k)\)得到的不同排列个数。

\(n\le 2\cdot 10^5,k\le 5\)

题解

一个置换相当于\(k!\)条从一个排列到另一个排列的有向边，由于肯定会成环，所以不用考虑方向性。

那么\(f(l,r)\)相当于l到r的边加上之后初始排列所在联通块的大小。

考虑枚举右端点，维护最大生成树（权值是置换的编号i)。

每个排列的贡献就是它到初始排列路径上边权的最大值。

由于最多\(k!\)个不同排列，所以每次维护复杂度\((k!)^2\)

实际上每次只需要用原来最大生成树和新边建树就可以了，\(O(nk!)\)

/*
Author: QAQ Automaton
Lang: C++
Prog: D.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	// input a signed integer
	inline bool read (signed &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}

	inline bool read (long long &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		x=gc();
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	// print a signed integer
	inline bool write (signed x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}

	inline bool write (long long x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int inf;
struct _init_{
    _init_(){
        memset(&inf,0x3f,sizeof(inf));
    }
};
_init_ ___INIT__;
int a[200005];
int to[125][125];
int w[6];
int is[3028];
int nx[125][6];
int fa[125],siz[125];
vector<pair<pii,int> >tree,ot;
int ans;
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void Link(int u,int v,int w){
	if(find(u)!=find(v)){
		if(find(u)==find(1)){ans+=siz[find(v)]*w;}
		if(find(v)==find(1)){ans+=siz[find(u)]*w;}
		if(siz[find(u)]>siz[find(v)])swap(u,v);
		siz[find(v)]+=siz[find(u)];
		fa[find(u)]=find(v);
		tree.push_back(make_pair(make_pair(u,v),w));
	}
}
signed main(){
#ifdef QAQAutoMaton 
	freopen("D.in","r",stdin);
	freopen("D.out","w",stdout);
#endif
	int n,k;
	read(n,k);
	for(int i=0;i<k;++i)w[i]=i;
	int t=0;
	do{
		int x=0;
		for(int i=0;i<k;++i)x=x*k+w[i];
		is[x]=++t;
		for(int i=0;i<k;++i)nx[t][i]=w[i];
	}while(next_permutation(w,w+k));
	for(int i=1;i<=t;++i)
		for(int j=1;j<=t;++j){
			int s=0;
			for(int l=0;l<k;++l)s=s*k+nx[i][nx[j][l]];
			to[i][j]=is[s];
		}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=k;++j){
			int x;
			read(x);
			a[i]=a[i]*k+(x-1);
		}
		a[i]=is[a[i]];
		for(int j=1;j<=t;++j){fa[j]=j;siz[j]=1;}
		ans+=i;
		ot=tree;
		tree.clear();
		for(int j=1;j<=t;++j)Link(j,to[j][a[i]],i);
		for(auto j:ot)Link(j.x.x,j.x.y,j.y);
	}
	write(ans,'\n');
	return 0;
}