CF1043D - Mysterious Crime

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 << "\n";
#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 '\n'

#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\n", ans);
    return 0;
}