Bzoj 1014&Luogu 4036 火星人Prefix(FHQ-Treap)

题面

洛谷

Bzoj

题解

首先,这种带修改的是不能用$SA$的,然后,我们做$SA$的题一般也能二分+$Hash$,所以不妨考虑用$FHQ-Treap$维护树,然后查询就用二分+$Hash$。

$Hash$怎么写?

$ hash[o]=hash[lc[o]]\times base[siz[rc[o]]+1]+val[o]\times base[siz[rc[o]]]+hash[rc[o]] $

为什么可以这么写呢?想一想,为什么

我们一般怎么求一颗维护序列的平衡树的原序列呢?—中序遍历

所以嘛,一棵子树的哈希值可以转化成它左子树的哈希值+本身的值+右子树哈希值

由于怕重复,所以可以考虑将前面两个值随便乘上一点什么东西(比如左子树或者右子树的$size$之类的)

#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

template<typename T>
void read(T &x) {
    int flag = 1; x = 0; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}

const int N = 3e5 + 10;
char s[N];
int m, len, base[N];
int rt, tot, lc[N], rc[N], pri[N], siz[N], h[N], val[N];

inline void upt(int o) {
    siz[o] = siz[lc[o]] + siz[rc[o]] + 1;
    h[o] = h[lc[o]] * base[siz[rc[o]] + 1] + val[o] * base[siz[rc[o]]] + h[rc[o]];
}
inline int node(int x) { siz[++tot] = 1, val[tot] = h[tot] = x, pri[tot] = rand(); return tot; }
void split(int o, int k, int &l, int &r) {
    if(!o) { l = r = 0; return ; }
    if(siz[lc[o]] < k) l = o, split(rc[o], k - siz[lc[o]] - 1, rc[o], r);
    else r = o, split(lc[o], k, l, lc[o]);
    upt(o);
}
int merge(int l, int r) {
    if(!l || !r) return l + r;
    if(pri[l] < pri[r]) { rc[l] = merge(rc[l], r), upt(l); return l; }
    else { lc[r] = merge(l, lc[r]), upt(r); return r; }
}
inline int query(int l, int r) {
    int x, y, k, ret;
    split(rt, r, x, y), split(x, l - 1, x, k);
    ret = h[k], rt = merge(merge(x, k), y);
    return ret;
}

int main () {
    scanf("%s", s + 1), len = strlen(s + 1), srand(19260817), base[0] = 1;
    for(int i = 1; i < N; ++i) base[i] = base[i - 1] * 27;
    for(int i = 1; i <= len; ++i) rt = merge(rt, node(s[i] - 'a' + 1));
    read(m); char opt[5], ch[5]; int x, y, l, r, k;
    while(m--) {
        scanf("%s", opt), read(x);
        if(opt[0] == 'Q') {
            read(y); int L = 0, R = std::min(len - x, len - y) + 1, ret;
            while(L <= R) {
                int mid = (L + R) >> 1;
                if(query(x, x + mid - 1) == query(y, y + mid - 1)) ret = mid, L = mid + 1;
                else R = mid - 1;
            } printf("%d\n", ret);
        } else {
            scanf("%s", ch), split(rt, x, l, r), ++len;
            if(opt[0] == 'R') --len, split(l, x - 1, l, k);
            rt = merge(merge(l, node(ch[0] - 'a' + 1)), r);
        }
    }
    return 0;
}
posted @ 2018-12-22 16:04  water_mi  阅读(184)  评论(0编辑  收藏  举报