大意: 给定m个n排列, 求有多少个公共子串.
枚举每个位置, hash求出最大匹配长度.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e5+10; int n, m, lst, s[12][N]; int pos[12][N]; int bas[N], has[12][N]; int Hash(int id, int l, int r) { int ret = (has[id][r]-(ll)has[id][l-1]*bas[r-l+1])%P; if (ret<0) ret += P; return ret; } int calc(int x, int p1, int y, int p2) { int r=min({lst+1,p1,p2}); return Hash(x,p1-r+1,p1)==Hash(y,p2-r+1,p2)?r:1; } int main() { scanf("%d%d", &n, &m); bas[0] = 1, bas[1] = n*20+1; REP(i,2,n) bas[i]=(ll)bas[i-1]*bas[1]%P; REP(i,1,m) REP(j,1,n) { scanf("%d", s[i]+j); has[i][j] = ((ll)has[i][j-1]*bas[1]+s[i][j])%P; pos[i][s[i][j]] = j; } if (m==1) return printf("%lld ",(ll)n*(n+1)/2),0; ll ans = 0; REP(i,1,n) { int d = 1e9; REP(j,2,m) d = min(d, calc(1,i,j,pos[j][s[1][i]])); ans += d; lst = d; } printf("%lld ", ans); }