luoguP4036 [JSOI2008]火星人 平衡树+hash

这个操作十分的复杂

但是可以拿平衡树维护

直接二分答案然后用$hash$值判断即可

复杂度$O(10000 * log^2 n + n \log n)$

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define ri register int
#define ull unsigned long long
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)
#define gc getchar
inline int read() {
    int p = , w = ; char c = gc();
    while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
    while(c >= '' && c <= '') p = p *  + c - '', c = gc();
    return p * w;
}

inline char gcr() {
    char c = gc();
    while() {
        if(c >= 'a' && c <= 'z') return c;
        if(c >= 'A' && c <= 'Z') return c;
        c = gc();
    }
}

const int sid = ;
const int sed = ;

char s[sid];
int n, m, rt, at, bt, ct, dt, id;
int sz[sid], ls[sid], rs[sid], pri[sid];
ull wei[sid], sum[sid], val[sid];

inline int rand() {
    static int seed = ;
    return seed = (1ll * seed * ) % ;
}

inline int newnode(char v) {
    ++ id;
    val[id] = v; sz[id] = ;
    pri[id] = rand(); sum[id] = v;
    return id;
}

inline void upd(int o) {
    sz[o] = sz[ls[o]] + sz[rs[o]] + ;
    sum[o] = sum[ls[o]] + sum[rs[o]] * wei[sz[ls[o]] + ];
    sum[o] += val[o] * wei[sz[ls[o]] + ];
}

inline int merge(int x, int y) {
    if(!x || !y) return x + y;
    if(pri[x] > pri[y]) {
        rs[x] = merge(rs[x], y);
        upd(x); return x;
    }
    else {
        ls[y] = merge(x, ls[y]);
        upd(y); return y;
    }
}

inline void split(int o, int k, int &x, int &y) {
    if(!o) { x = y = ; return; }
    if(k <= sz[ls[o]]) y = o, split(ls[o], k, x, ls[o]);
    else x = o, split(rs[o], k - sz[ls[o]] - , rs[o], y);
    upd(o);
}

inline ull getHash(int l, int r) {
    split(rt, r, at, bt);
    split(at, l - , ct, dt);
    ull ret = sum[dt];
    rt = merge(ct, merge(dt, bt));
    return ret;
}

inline bool check(int x, int y, int v) {
    return getHash(x, x + v - ) == getHash(y, y + v - );
}

int main() {
    scanf("%s", s + );
    n = strlen(s + );

    wei[] = ;
    rep(i, , ) wei[i] = wei[i - ] * sed;
    rep(i, , n) rt = merge(rt, newnode(s[i]));

    m = read();
    rep(i, , m) {
        char c = gcr();
        int x, y; char d;
        if(c == 'Q') {
            x = read(); y = read();
            if(x > y) swap(x, y);
            int l = , r = n - y + , ans = ;
            while(l <= r) {
                int mid = (l + r) >> ;
                if(check(x, y, mid)) {
                    l = mid + , ans = mid;
                    printf("qaq -> %d\n", mid);
                }
                else r = mid - ;
            }
            printf("%d\n", ans);
        }
        else if(c == 'R') {
            x = read(); d = gcr();
            split(rt, x - , at, bt);
            split(bt, , ct, dt);
            rt = merge(at, merge(newnode(d), dt));
        }
        else if(c == 'I') {
            n ++;
            x = read(); d = gcr();
            split(rt, x, at, bt);
            rt = merge(at, merge(newnode(d), bt));
        }
    }
    return ;
}