LightOJ 1268 Unlucky Strings (KMP+矩阵快速幂)

时间：2024-08-19 12:36:20

题意：给出一个字符集和一个字符串和正整数n，问由给定字符集组成的所有长度为n的串中不以给定字符串为连续子串的有多少个？

析：n 实在是太大了，如果小的话，就可以用动态规划做了，所以只能用矩阵快速幂来做了，dp[i][j] 表示匹配完 i 到匹配 j 个有多少种方案，利用矩阵的性质，就可以快速求出长度为 n 的个数，对于匹配的转移，正好可以用KMP的失配函数来转移。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-6;

const int maxn = 50 + 10;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r > 0 && r <= n && c > 0 && c <= m;

}

char s[maxn], t[maxn];

int f[maxn];

struct Matrix{

  unsigned int a[maxn][maxn];

  int n;

  Matrix(int nn) : n(nn){

    memset(a, 0, sizeof a);

  }

  void toOne(){

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

      a[i][i] = 1;

  }

  friend Matrix operator * (const Matrix &lhs, const Matrix &rhs){

    Matrix res(lhs.n);

    for(int i = 0; i < lhs.n; ++i)

      for(int j = 0; j < rhs.n; ++j)

        for(int k = 0; k < lhs.n; ++k)

          res.a[i][j] += lhs.a[i][k] * rhs.a[k][j];

    return res;

  }

};

Matrix fast_pow(Matrix a, int n){

  Matrix res(a.n);

  res.toOne();

  while(n){

    if(n&1) res = res * a;

    a = a * a;

    n >>= 1;

  }

  return res;

}

void getFail(char *s, int n){

  f[0] = f[1] = 0;

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

    int j = f[i];

    while(j && s[j] != s[i])  j = f[j];

    f[i+1] = s[i] == s[j] ? j+1 : 0;

  }

}

int main(){

  int T;  cin >> T;

  for(int kase = 1; kase <= T; ++kase){

    scanf("%d", &n);

    scanf("%s %s", t, s);

    int lens = strlen(s);

    getFail(s, lens);

    Matrix ans(lens);

    for(int i = 0; i < lens; ++i)

      for(int k = 0; t[k]; ++k){

        int j = i;

        while(j && s[j] != t[k])  j = f[j];

        if(s[j] == t[k])  ++j;

        ++ans.a[i][j];

      }

    ans = fast_pow(ans, n);

    unsigned int res = 0;

    for(int i = 0; i < lens; ++i)  res += ans.a[0][i];

    printf("Case %d: %u\n", kase, res);

  }

  return 0;

}

　　

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