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,$O(nk!)$

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/*
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;
}

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