c/c++语言开发共享洛谷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("%dn", 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 */