KMP暴力求出next数组后
实际上是一个最短路问题，floyed搞一搞
然而会TLE
矩阵优化一下即可（倍增floyed）
KMP在弱数据下可以AC。。正解请看其他人博客

# include <stdio.h>

# include <stdlib.h>

# include <iostream>

# include <algorithm>

# include <string.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

IL ll Read(){

    RG char c = getchar(); RG ll x = 0, z = 1;

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';

    return x * z;

}

int n, m, nxt[210][100010], len[210];

char s[210][100010];

struct Matrix{

    ll a[210][210];

    IL void Clear(){  Fill(a, 63);  }

    IL ll* operator [](RG int x){  return a[x];  }

    IL Matrix operator *(RG Matrix &B){

        RG Matrix C; C.Clear();

        for(RG int i = 0; i <= n; i++)

            for(RG int j = 0; j <= n; j++)

                for(RG int k = 0; k <= n; k++)

                    C[i][k] = min(C[i][k], a[i][j] + B[j][k]);

        return C;

    }

} ans, edge;

int main(RG int argc, RG char* argv[]){

    edge.Clear(); ans.Clear();

    n = Read(); m = Read();

    for(RG int i = 1; i <= n; i++)

        scanf(" %s", s[i] + 1), len[i] = strlen(s[i] + 1);

    for(RG int k = 1; k <= n; k++)

        for(RG int i = 2, j = 0; i <= len[k]; i++){

            while(j && s[k][j + 1] != s[k][i]) j = nxt[k][j];

            if(s[k][j + 1] == s[k][i]) j++;

            nxt[k][i] = j;

        }

    for(RG int x = 1; x <= n; x++){

        edge[0][x] = len[x];

        for(RG int y = 1; y <= n; y++)

            for(RG int i = 2, j = 0; i <= len[x]; i++){

                while(j && s[y][j + 1] != s[x][i]) j = nxt[y][j];

                if(s[y][j + 1] == s[x][i]) j++;

                if(i == len[x]) edge[x][y] = len[y] - j;

            }

    }

    for(RG int i = 0; i <= n; i++) ans[i][i] = 0;

    for(; m; m >>= 1, edge = edge * edge) if(m & 1) ans = ans * edge;

    RG ll min_len = 1e18;

    for(RG int i = 1; i <= n; i++) min_len = min(ans[0][i], min_len);

    printf("%lld\n", min_len);

    return 0;

}