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;
}
/*

*/