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
*/