luogu P5310 [Ynoi2011] 遥远的过去

https://www.luogu.com.cn/problem/P5310

对于这种要离散化，再匹配的一般的字符串算法都不好处理
考虑哈希
显然可以在splay上维护每个位置都hash值

$ha=\sum pos[i]\ast B^{rank[i]}$

对于A每个长度为m的子串都把哈希值求出来扔到哈希表里
然后再拿splay维护B的哈希值即可
直接在哈希表里找
code:

#include<bits/stdc++.h>
#define N 200050
#define ls ch[x][0]
#define rs ch[x][1]
#define ll unsigned long long
using namespace std;
int size[N], ch[N][2], fa[N], root, v[N], n, m, q;
ll ha[N], mi[N], S;
const ll B = 11117;
const ll mod = 100003;
void update(int x) {
    size[x] = size[ls] + size[rs] + 1;
    ha[x] = ha[ls] + mi[size[ls]] * (ll)x + mi[size[ls] + 1] * ha[rs];
}
int get(int x) {
    return ch[fa[x]][1] == x;
}
void rotate(int x) {
    int f = fa[x], gf = fa[f], k = get(x);
    if(!f) return ;
    if(gf) ch[gf][get(f)] = x; fa[x] = gf;
    ch[f][k] = ch[x][!k], fa[ch[x][!k]] = f;
    ch[x][!k] = f, fa[f] = x;
    update(f), update(x);
}
inline void splay(int x) {
    int f = fa[x];
    while(f) {
        if(fa[f]) rotate(get(x) == get(f)? f : x);
        rotate(x);
        f = fa[x];
       // printf("%d %d\n", x, f);
    }
    root = x;
}
void insert(int &x, int id, int f) {
   // printf("%d %d %d   \n", x, id, f);
    if(!x) {x = id; fa[x] = f; update(x); splay(x); return ;}
    if(v[id] < v[x]) insert(ls, id, x);
    else insert(rs, id, x);
}
void clear(int x) {
    fa[x] = ls = rs = size[x] = 0;
}
void del(int x) {
    splay(x);
    if(!ls || !rs) {root = ls + rs; fa[root] = 0; clear(x); return ; }
    int now = ls;
    while(ch[now][1]) now = ch[now][1];
    splay(now);
    ch[now][1] = ch[x][1], fa[ch[x][1]] = now;
    clear(x); update(now);
}
void deb(int x) {
    if(!x) return ;
    deb(ls);
    printf(" %d %d %d   ", x, ls, rs);
    deb(rs);
}


struct HAT {
    int nxt, s;
    ll val;
} e[N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void INS(ll x) {
    int u = x % mod;
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        if(x == e[i].val) { ++ e[i].s; return ;}
    }
    e[eid].val = x, e[eid].s = 1, e[eid].nxt = p[u];
    p[u] = eid ++;
}
int QRY(ll x) {
    int u = x % mod;
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        if(x == e[i].val)  return e[i].s;
    } return 0;
}
int main() {
    init();
    scanf("%d%d%d", &n ,&m, &q);
    mi[0] = 1;
    for(int i = 1; i <= n; i ++) mi[i] = mi[i - 1] * B;
    
    for(int i = 1; i <= n; i ++) scanf("%d", &v[i]);
    
    for(int i = 1; i <= m; i ++) insert(root, i, 0), S += mi[i - 1];
    for(int i = m; i <= n; i ++) {
        INS(ha[root] - S * (i - m));
      //  printf("\n **  %llu   ", ha[root]); deb(root); printf("\n");
        del(i - m + 1);
     //   printf("\n **  "); deb(root); printf("\n");
        insert(root, i + 1, 0);
    }
    root = 0;
    
    memset(ch, 0, sizeof ch), memset(fa, 0, sizeof fa);
    for(int i = 1; i <= m; i ++) scanf("%d", &v[i]), insert(root, i, 0);
    while(q --) {
        int x, y;
        scanf("%d%d", &x, &y);
        del(x);
        v[x] = y; insert(root, x, 0);
        printf("%d\n", QRY(ha[root]));
    }
    return 0;
}