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

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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