cf1043D. Mysterious Crime(枚举)

题意

题目链接

给出(m)个长度为(n)的排列，问有多少连续公共子串

(m leqslant 10, n leqslant 10^5)

Sol

非常naive的一道题然而交了3遍才过(昨晚真的困得不行。。)

枚举第一个串的位置，直接维护10个指针，分别表示每个串与第一个串匹配到的位置，同时记录出已经匹配了多少位

因为题目中给出的是排列，所以如果某一位不能匹配了，则需要从新的位置开始

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e5 + 10, INF = 1e9 + 10, mod = 998244353;
const double eps = 1e-9;
inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, M;
int a[11][MAXN], pos[11][MAXN], now[MAXN], len[MAXN];
main() {
	N = read(); M = read();
	for(int i = 1; i <= M; i++)
		for(int j = 1; j <= N; j++) {
			a[i][j] = read();
			pos[i][a[i][j]] = j;
		}
	LL ans = 0;
	for(int i = 1; i <= N; i++) {
		len[1] = i;
		for(int j = 2; j <= M; j++) {
			if(a[j][now[j] + 1] != a[1][i]) now[j] = pos[j][a[1][i]], len[j] = 1;
			else now[j]++, len[j]++;
		}
		LL mn = INF;
		for(int j = 1; j <= M; j++) mn = min(mn, (LL)len[j]);
		ans += mn;
	}
	cout << ans;
	return 0;
}

/*
*/