Foj 2299 Prefix(AC自动机、DP)
Foj 2299 Prefix
题意
给定串s、正整数n,问有多少长度为n的字符串t满足:s[0...i]是t的子串,s[0...i+1]不是。
题解
求有多少长度为n的字符串t满足:s[0...i]是t的子串
这样求出来的数列就是答案的一个前缀和。
具体求法可以对于s[0...i]建一个AC自动机,列出dp式子,状态转移矩阵就很好构造了。
现场主要是没想到每个前缀的方案数要分开求。
代码
#include<cstdio>
#include<cmath>
#include<vector>
#include<iostream>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const int N = 22, P = 1e9+7;
int n, m;
int f[N];
string s;
int kpow(int a, int b) {
int res = 1;
while(b) {
if(b&1) res = 1ll*res*a%P;
a = 1ll*a*a%P;
b >>= 1;
}
return res;
}
inline int add(int a, int b) {
a+=b;
if(a>=P) a-=P;
return a;
}
inline int sub(int a, int b) {
a-=b;
if(a<0) a+=P;
return a;
}
inline int mul(int a, int b) {
return 1ll*a*b%P;
}
struct Mat {
static const int N = ::N;
int a[N][N], n;
Mat(){} Mat(int _n, int v) { n = _n; rep(i, 0, n) rep(j, 0, n) a[i][j] = i==j ? v : 0; }
Mat operator * (const Mat &c) const {
Mat res(n, 0);
rep(i, 0, n) rep(j, 0, n) rep(k, 0, n) res.a[i][j] = add(res.a[i][j], mul(a[i][k], c.a[k][j]));
return res;
}
Mat operator ^ (int b) const {
Mat res(n, 1), a = *this;
while(b) {
if(b&1) res = res*a;
a = a*a;
b>>=1;
}
return res;
}
};
struct Trie {
static const int N = ::N, M = 26;
int ne[N][M], fail[N], fa[N], rt, L, ed[N];
void init() {
fill_n(ne[fail[0] = N-1], M, 0);
fill_n(ed, L, 0);
L = 0;
rt = newnode();
}
int newnode() {
fill_n(ne[L], M, 0);
return L++;
}
void add(string s) {
int p = rt;
rep(i, 0, sz(s)) {
int c = s[i] - 'A';
if(!ne[p][c]) ne[p][c] = newnode(), fa[L-1] = p;
p = ne[p][c];
}
ed[p] = 1;
}
void build() {
vi v;v.pb(rt);
rep(i, 0, sz(v)) {
int c = v[i];
rep(i, 0, M) ne[c][i] ?
v.pb(ne[c][i]), fail[ne[c][i]] = ne[fail[c]][i] :
ne[c][i] = ne[fail[c]][i];
}
}
Mat getMat() {
Mat mat(L, 0);
rep(i, 0, L) rep(j, 0, M) {
int nex = ed[i] ? i : ne[i][j];
++mat.a[i][nex];
}
return mat;
}
}ac;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int T;
cin >> T;
while(T--) {
cin >> n >> m >> s;
f[0] = kpow(26, n);
rep(i, 1, m+1) {
ac.init();
ac.add(s.substr(0, i));
ac.build();
Mat res = ac.getMat() ^ n;
f[i] = res.a[0][i];
}
rep(i, 0, m) f[i] = sub(f[i], f[i+1]);
rep(i, 0, m+1) cout << f[i] << endl;
}
return 0;
}