题解 P9695【[GDCPC2023] Traveling in Cells】

显然，询问的答案即为 $x$ 所在的极长的满足颜色均在 $\mathbb{A}$ 内的连续段的权值和。如果我们能维护对颜色的单点修改，以及求出某个位置所在极长连续段的左右端点 $l,r$，只需要树状数组即可求出答案。

一个朴素的想法是对每种颜色开一棵线段树，单点修改是平凡的，极长连续段左端点只需要线段树上二分找到最靠右的 $l$，使得 $\mathbb{A}$ 中每种颜色在 $[l,x]$ 的出现次数之和恰好等于 $x-l+1$，极长连续段右端点同理。

咱赛时没算这玩意的空间开销有多大，就直接莽了个主席树上去。主席树的话空间复杂度显然为 $O((n+q)\log n)$，实现中开了 $64$ 倍数组。

赛时懒得写线段树上二分的 $O((n+q)\log n)$ 做法，写了二分套线段树的 $O(n\log n+q\log^2n)$ 跑过去了。

// Problem: T368285 [GDCPC2023] F-Traveling in Cells
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/T368285?contestId=135929
// Memory Limit: 1 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(ll x = (y); x <= (z); ++x)
#define per(x, y, z) for(ll x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
ll randint(ll L, ll R) {
    uniform_int_distribution<ll> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

const ll N = 1e5 + 5;

ll T, n, q, c[N], v[N];

struct BIT {
    ll c[N];
    void init(ll x) {rep(i, 1, x) c[i] = 0;}
    ll lowbit(ll x) {return x & (-x);}
    void add(ll x, ll k) {for(; x <= n; x += lowbit(x)) c[x] += k;}
    ll ask(ll x) {ll k = 0; for(; x; x -= lowbit(x)) k += c[x]; return k;}
    ll Ask(ll l, ll r) {return ask(r) - ask(l - 1);}
}bit;

struct PersSegTree {
    ll cnt[N << 6], lc[N << 6], rc[N << 6], rt[N], sz;
    void clear() {
        rep(i, 1, sz) cnt[i] = lc[i] = rc[i] = 0;
        rep(i, 1, n) rt[i] = 0;
        sz = 0;
    }
    ll copy(ll u) {
        ++sz;
        cnt[sz] = cnt[u];
        lc[sz] = lc[u];
        rc[sz] = rc[u];
        return sz;
    }
    void pushup(ll u) {
        cnt[u] = cnt[lc[u]] + cnt[rc[u]];
    }
    ll insert(ll u, ll l, ll r, ll pos, ll k) {
        ll v = copy(u);
        if(l == r) {
            cnt[v] += k;
            return v;
        }
        ll mid = (l + r) >> 1;
        if(pos <= mid) lc[v] = insert(lc[u], l, mid, pos, k);
        else rc[v] = insert(rc[u], mid + 1, r, pos, k);
        pushup(v);
        return v;
    }
    ll query(ll u, ll l, ll r, ll ql, ll qr) {
        if(!u) return 0;
        if(ql <= l && r <= qr) return cnt[u];
        ll mid = (l + r) >> 1, ans = 0;
        if(ql <= mid) ans += query(lc[u], l, mid, ql, qr);
        if(qr > mid) ans += query(rc[u], mid + 1, r, ql, qr);
        return ans;
    }
}psgt;

ll findL(const vector<ll>& vec, ll x) {
    ll L = 1, R = x + 1;
    auto check = [&](ll M) -> bool {
        ll len = x - M + 1, cnt = 0;
        for(ll c : vec) cnt += psgt.query(psgt.rt[c], 1, n, M, x);
        return cnt == len;
    };
    while(L < R) {
        ll M = (L + R) >> 1;
        if(check(M)) R = M;
        else L = M + 1;
    }
    return L;
}

ll findR(const vector<ll>& vec, ll x) {
    ll L = x, R = n + 1;
    auto check = [&](ll M) -> bool {
        ll len = M - x + 1, cnt = 0;
        for(ll c : vec) cnt += psgt.query(psgt.rt[c], 1, n, x, M);
        return cnt == len;
    };
    while(L < R) {
        ll M = (L + R) >> 1;
        if(check(M)) L = M + 1;
        else R = M;
    }
    return L - 1;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    for(cin >> T; T; --T) {
        cin >> n >> q;
        rep(i, 1, n) cin >> c[i];
        rep(i, 1, n) cin >> v[i];
        bit.init(n);
        psgt.clear();
        rep(i, 1, n) bit.add(i, v[i]);
        rep(i, 1, n) psgt.rt[c[i]] = psgt.insert(psgt.rt[c[i]], 1, n, i, +1);
        while(q--) {
            ll op;
            cin >> op;
            if(op == 1) {
                ll x, y;
                cin >> x >> y;
                psgt.rt[c[x]] = psgt.insert(psgt.rt[c[x]], 1, n, x, -1);
                c[x] = y;
                psgt.rt[c[x]] = psgt.insert(psgt.rt[c[x]], 1, n, x, +1);
            }
            else if(op == 2) {
                ll x, y;
                cin >> x >> y;
                bit.add(x, -v[x]);
                v[x] = y;
                bit.add(x, +v[x]);
            }
            else {
                ll x, k;
                cin >> x >> k;
                vector<ll> vec(k);
                for(ll& i : vec) cin >> i;
                ll L = findL(vec, x), R = findR(vec, x);
                cout << bit.Ask(L, R) << endl;
            }
        }
    }
    return 0;
}