洛谷P4591 [TJOI2018]碱基序列(hash dp)

题意

题目链接

Sol

\(f[i][j]\)表示匹配到第\(i\)个串，当前在主串的第\(j\)个位置

转移的时候判断一下是否可行就行了。随便一个能搞字符串匹配的算法都能过

复杂度\(O(|S| K a_i)\)

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define ull signed long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 3e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
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 K;
char s[MAXN], tmp[MAXN];
int f[101][100001];
ull ha[MAXN], po[MAXN], base = 27;
ull Query(int l,int r) {
	return ha[r] - po[r - l + 1] * ha[l - 1];
}
signed main() {
	K = read();
	scanf("%s", s + 1);
	int N = strlen(s + 1); po[0] = 1;
	for(int i = 1; i <= N; i++) po[i] = base * po[i - 1], ha[i] = ha[i - 1] * base + s[i];
	for(int i = 0; i <= N; i++) f[0][i] = 1;
	for(int i = 1; i <= K; i++) {
		int num = read();
		for(int j = 1; j <= num; j++) {
			scanf("%s", tmp + 1);
			int l = strlen(tmp + 1);
			ull val = 0;
			for(int k = 1; k <= l; k++) val = val * base + tmp[k];
			for(int k = l; k <= N; k++) {
				if(val == Query(k - l + 1, k)) {
					add2(f[i][k], f[i - 1][k - l]);
				}
			}
		}
	}
	int ans = 0;
	for(int i = 1; i <= N; i++) add2(ans, f[K][i]);
	cout << ans;
    return 0;
}
/*

*/