CF1043D

CF1043D

这道题也不能说是dp，感觉dp没那么强。

题目是说，给定m（m<=10）个数列，每个数列包含n个数（n<=1e5).问这m个数列中有多少个相同的子数列（连续的哦）

重点在于透过第一个数列，即以第一个数列为样本，找到符合条件的子数列。从第一个数列的最后开始向前遍历，每次判断m条数列是不是都满足第i个位子的数字x的后面是数字y。如果不是，这个位置dp值记为1，否则就是dp【i+1】+1。答案加上每个位子的dp值即可。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x7f7f7f7f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e8+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 1e5+9;
            ll mp[20][maxn],a[20][maxn],dp[maxn];
int main(){
            int n,m;
            scanf("%d%d", &n, &m);
            for(int i=1; i<=m; i++){
                for(int j=1; j<=n; j++){
                    scanf("%lld", &mp[i][j]);
                    a[i][mp[i][j]] = j;
                }
            }   
            ll ans = 0;
            for(int i=n; i>=1; i--){
                int flag = 1, val = mp[1][i];
                dp[a[1][val]] = 1;
                for(int j=2; j<=m; j++){
                    if(a[j][val]+1>n || a[1][val]+1>n || mp[j][a[j][val]+1] != mp[1][a[1][val]+1]){
                        flag = 0;
                    }
                }
                if(flag)dp[i] = dp[i+1] + 1;
                ans += dp[i];
            }
            printf("%lld
", ans);
            return 0;
}