洛谷P3763 [TJOI2017]DNA(后缀数组 RMQ)

题意

题目链接

Sol

这题打死我也不会想到后缀数组的，应该会全程想AC自动机之类的吧

但知道这题能用后缀数组做之后应该就不是那么难了

首先把\(S\)和\(S0\)拼到一起跑，求出Height数组

暴力枚举每个后缀是否能成为答案。

具体来说，每次比较当前后缀和\(S_0\)的lcp，如果长度\(< N\)的话就从不合法的位置继续匹配

rmq维护一下区间lcp最小值

BZOJ上被完美卡常

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
const int INF = 2333;
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 N, M, L, rak[MAXN], tax[MAXN], tp[MAXN], sa[MAXN], H[MAXN], f[MAXN][20], lg2[MAXN];
char s[MAXN], s0[MAXN];
void Qsort() {
    for(int i = 0; i <= M; i++) tax[i] = 0;
    for(int i = 1; i <= N; i++) tax[rak[i]]++;
    for(int i = 1; i <= M; i++) tax[i] += tax[i - 1];
    for(int i = N; i >= 1; i--) sa[tax[rak[tp[i]]]--] = tp[i];
}
void SuffixSort() {
    for(int i = 1; i <= N; i++) rak[i] = s[i], tp[i] = i; M = 233; Qsort();
    for(int w = 1, p = 0; p < N; w <<= 1, M = p) { p = 0;
        for(int i = 1; i <= w; i++) tp[++p] = N - i + 1;
        for(int i = 1; i <= N; i++) if(sa[i] > w) tp[++p] = sa[i] - w;
        Qsort(); swap(tp, rak); rak[sa[1]] = p = 1;
        for(int i = 2; i <= N; i++) rak[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w]) ? p : ++p;
    }
    for(int i = 1, k = 0; i <= N; i++) {
        if(k) k--; int j = sa[rak[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        H[rak[i]] = k;
    }
}
void Pre() {
    for(int i = 1; i <= N; i++) f[i][0] = H[i];
    for(int j = 1; j <= 17; j++)
        for(int i = 1; i + (1 << j) - 1 <= N; i++) f[i][j] = min(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
}
int Query(int x, int y) {
    if(x > y) swap(x, y); x++;
    int k = lg2[y - x + 1];
    return min(f[x][k], f[y - (1 << k) + 1][k]);
}
int check(int x, int y, int dep) {
    if(dep == 3) return Query(rak[x], rak[y]);
    int num = Query(rak[x], rak[y]);
    num +=  check(x + num + 1, y + num + 1, dep + 1) + 1;
    return num;
}
void solve() {
    scanf("%s%s", s + 1, s0 + 1);
    L = strlen(s0 + 1); N = strlen(s + 1);
    for(int i = 1; i <= L; i++) s[N + i] = s0[i];
    N += L;
    SuffixSort(); Pre(); N -= L; int ans = 0;
    for(int i = 1; i <= N - L + 1; i++) if(check(i, N + 1, 0) >= L) ans++;
    printf("%d\n", ans);
}
int main() {
    //freopen("a.in", "r", stdin);
    lg2[1] = 0; for(int i = 2; i <= MAXN - 1; i++) lg2[i] = lg2[i >> 1] + 1;
    for(int T = read(); T; solve(), T--);
    return 0;
}
/*
2
ATCGCCCTA
CTTCA
ATCGCCCTA
CTTCA
*/