SCUT

https://scut.online/p/106

错在这组样例，发现是离散化之后，对k访问的时候也是应该访问离散化之后的k。

12 4
1 1 2 2 5 5 4 4 3 3 2
1 1 3 3 5 7 7 9 9 9 11 11
1 10
3 10
3 11
2 4

发现主席树大概还真的要开够log倍，少一点都不行，那干脆开大一点。

#include<bits/stdc++.h>
#define mid ((l+r)>>1)
using namespace std;

const int MAXN = 100000 + 5;

int a[MAXN], b[MAXN];
vector<int> E[MAXN];

int siz[MAXN], tid[MAXN], rnk[MAXN], cnt;

int T[MAXN], tcnt;
int sum[MAXN << 5], L[MAXN << 5], R[MAXN << 5];

void init(int n) {
    for(int i = 1; i <= n; ++i)
        E[i].clear();
    cnt = 0;
    tcnt = 0;
}

void dfs(int u) {
    siz[u] = 1;
    tid[u] = ++cnt;
    rnk[cnt] = u;
    for(auto v : E[u]) {
        dfs(v);
        siz[u] += siz[v];
    }
}

inline int build(int l, int r) {
    int rt = ++tcnt;
    sum[rt] = 0;
    if(l < r) {
        L[rt] = build(l, mid);
        R[rt] = build(mid + 1, r);
    }
    return rt;
}

inline int update(int pre, int l, int r, int x) {
    int rt = ++tcnt;
    R[rt] = R[pre];
    L[rt] = L[pre];
    sum[rt] = sum[pre] + 1;
    if(l < r) {
        if(x <= mid)
            L[rt] = update(L[pre], l, mid, x);
        else
            R[rt] = update(R[pre], mid + 1, r, x);
    }
    return rt;
}

//查询[u-1,v]中不超过k的数的个数
inline int query2(int u, int v, int l, int r, int k) {
    int res = 0;
    while(l < r && k < r) {
        if(k >= mid) {
            res += sum[L[v]] - sum[L[u]];
            u = R[u], v = R[v], l = mid + 1;
        } else
            u = L[u], v = L[v], r = mid;
    }
    return res + (k >= l ? sum[v] - sum[u] : 0);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, q, r = 1;
    while(~scanf("%d%d", &n, &q)) {
        init(n);
        for(int i = 2, f; i <= n; ++i) {
            scanf("%d", &f);
            E[f].push_back(i);
        }
        dfs(r);
        for(int i = 1; i <= n; i ++) {
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(b + 1, b + 1 + n);
        int cb = unique(b + 1, b + 1 + n) - b - 1;
        tcnt = 0;
        T[0] = build(1, cb);
        for(int i = 1; i <= n; i ++) {
            int t = lower_bound(b + 1, b + 1 + cb, a[rnk[i]]) - b;
            T[i] = update(T[i - 1], 1, cb, t);
        }
        while(q--) {
            int u, k;
            scanf("%d%d", &u, &k);
            k = upper_bound(b + 1, b + 1 + cb, k) - b - 1;
            int l = tid[u], r = tid[u] + siz[u] - 1;
            printf("%d
", query2(T[l - 1], T[r], 1, cb, k));
        }
    }
    return 0;
}

可以启发式合并Treap。我们知道启发式合并的复杂度确实总共是nlogn级别的。离线所有操作，从叶子向上合并。那么每次询问的时候的确就是在u所在子树查询，因为u节点子树恰好就是u节点的平衡树。还是在方面的万能样例翻车了。是批量插入和删除有点问题。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 120000;

int ch[MAXN + 5][2];
int val[MAXN + 5], dat[MAXN + 5];
int siz[MAXN + 5], cnt[MAXN + 5];
ll sum[MAXN + 5];
int tot, root[MAXN + 5];

int st[MAXN + 5], stop;

inline void Init(int n) {
    tot = 0;
    memset(root, 0, sizeof(root[0]) * (n + 1));
    stop = 0;
}

inline int NewNode(int v) {
    if(stop == 0) {
        ch[++tot][0] = 0;
        ch[tot][1] = 0;
        val[tot] = v, dat[tot] = rand();
        siz[tot] = 1, cnt[tot] = 1;
        sum[tot] = v;
        return tot;
    } else {
        int id = st[stop--];
        ch[id][0] = 0;
        ch[id][1] = 0;
        val[id] = v, dat[id] = rand();
        siz[id] = 1, cnt[id] = 1;
        sum[id] = v;
        return id;
    }
}

inline void PushUp(int id) {
    siz[id] = siz[ch[id][0]] + siz[ch[id][1]] + cnt[id];
    sum[id] = sum[ch[id][0]] + sum[ch[id][1]] + 1ll * val[id] * cnt[id];
}

inline void Rotate(int &id, int d) {
    int temp = ch[id][d ^ 1];
    ch[id][d ^ 1] = ch[temp][d];
    ch[temp][d] = id;
    id = temp;
    PushUp(ch[id][d]), PushUp(id);
}

inline void Insert(int &id, int v) {
    if(!id)
        id = NewNode(v);
    else {
        if(v == val[id])
            ++cnt[id];
        else {
            int d = v < val[id] ? 0 : 1;
            Insert(ch[id][d], v);
            if(dat[id] < dat[ch[id][d]])
                Rotate(id, d ^ 1);
        }
        PushUp(id);
    }
}

void Remove(int &id, int v) {
    if(!id)
        return;
    else {
        if(v == val[id]) {
            if(cnt[id] > 1) {
                cnt[id] --;
                PushUp(id);
            } else if(ch[id][0] || ch[id][1]) {
                if(!ch[id][1] || dat[ch[id][0]] > dat[ch[id][1]])
                    Rotate(id, 1), Remove(ch[id][1], v);
                else
                    Rotate(id, 0), Remove(ch[id][0], v);
                PushUp(id);
            } else{
                st[++stop]=id;
                id = 0;
            }
        } else {
            v < val[id] ? Remove(ch[id][0], v) : Remove(ch[id][1], v);
            PushUp(id);
        }
    }
}

int GetRank(int id, int v) {
    //小于等于v的有几个？
    if(!id)
        return 0;
    else {
        if(v == val[id])
            return siz[ch[id][0]] + cnt[id];
        else if(v < val[id])
            return GetRank(ch[id][0], v);
        else
            return siz[ch[id][0]] + cnt[id] + GetRank(ch[id][1], v);
    }
}

//把id2的树整棵插进id1中
void Merge(int &id1, int &id2) {
    if(siz[id1] < siz[id2])
        swap(id1, id2);
    //将id2的根插进id1，目前每次只搬运一个
    while(siz[id2]) {
        int tmpv = val[id2]/*, tmpn = cnt[id2]*/;
        Remove(id2, tmpv);
        Insert(id1, tmpv);
    }
    return;
}

int d[MAXN], f[MAXN], a[MAXN];

int u2topo[MAXN], cnttopo, topo2u[MAXN];

struct Query {
    int u, k, ans, id;
    bool operator<(const Query& q)const {
        return u2topo[u] < u2topo[q.u];
    }
} que[MAXN];

struct cmp {
    bool operator()(const Query& q1, const Query& q2)const {
        return q1.id < q2.id;
    }
};

int que2[MAXN];
int front, back;

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, q;
    while(~scanf("%d%d", &n, &q)) {
        memset(d, 0, sizeof(d[0]) * (n + 1));
        for(int i = 2; i <= n; ++i) {
            scanf("%d", &f[i]);
            ++d[f[i]];
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        front = 1, back = 0, cnttopo = 0;
        for(int i = 1; i <= n; ++i) {
            if(d[i] == 0)
                que2[++back] = i;
        }
        while(back >= front) {
            int u = que2[front++];
            u2topo[u] = ++cnttopo;
            topo2u[cnttopo] = u;
            d[f[u]]--;
            if(d[f[u]] == 0)
                que2[++back] = f[u];
        }
        for(int i = 1; i <= q; ++i) {
            scanf("%d%d", &que[i].u, &que[i].k);
            que[i].id = i;
        }
        sort(que + 1, que + 1 + q);
        Init(n);
        for(int i = 1; i <= n; ++i) {
            Insert(root[i], a[i]);
            //cout<<siz[root[i]]<<endl;
            //cout<<u2topo[i]<<" ";
        }
        //cout<<endl;
        int curtopo = 1;
        for(int i = 1; i <= q; ++i) {
            while(u2topo[que[i].u] > curtopo) {
                //cout<<"Qu="<<que[i].u<<" topoQu="<<u2topo[que[i].u]<<endl;
                //等于的话不用合并，直接查询
                int u = topo2u[curtopo++];
                //cout<<"Merge "<<u<<" "<<f[u]<<endl;
                Merge(root[f[u]], root[u]);
                //cout<<"  After Merge Size="<<siz[root[f[u]]]<<endl;
            }
            //现在相等了
            que[i].ans = GetRank(root[que[i].u], que[i].k);
            /*cout<<"Answer u="<<que[i].u;
            cout<<" ans="<<que[i].ans<<endl;*/
        }
        sort(que + 1, que + 1 + q, cmp());
        for(int i = 1; i <= q; ++i) {
            printf("%d
", que[i].ans);
        }
        //puts("---");
    }
    return 0;
}

是NewNode没有传入num的问题，但是为什么恰好开n个位置会RE了呢？（因为后面的MAXN没有加5！以后还是在MAXN里面+5，这样数组会好看一点）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 120000;

int ch[MAXN + 5][2];
int val[MAXN + 5], dat[MAXN + 5];
int siz[MAXN + 5], cnt[MAXN + 5];
int tot, root[MAXN + 5];

int st[MAXN + 5], stop;

inline void Init(int n) {
    tot = 0;
    memset(root, 0, sizeof(root[0]) * (n + 1));
    stop = 0;
}

inline int NewNode(int v, int num) {
    int id = ((stop==0) ? ++tot : st[stop--]);
    ch[id][0] = 0;
    ch[id][1] = 0;
    val[id] = v, dat[id] = rand();
    siz[id] = num, cnt[id] = num;
    return id;
}

inline void PushUp(int id) {
    siz[id] = siz[ch[id][0]] + siz[ch[id][1]] + cnt[id];
}

inline void Rotate(int &id, int d) {
    int temp = ch[id][d ^ 1];
    ch[id][d ^ 1] = ch[temp][d];
    ch[temp][d] = id;
    id = temp;
    PushUp(ch[id][d]), PushUp(id);
}

inline void Insert(int &id, int v, int num) {
    if(!id)
        id = NewNode(v, num);
    else {
        if(v == val[id])
            cnt[id] += num;
        else {
            int d = v < val[id] ? 0 : 1;
            Insert(ch[id][d], v, num);
            if(dat[id] < dat[ch[id][d]])
                Rotate(id, d ^ 1);
        }
        PushUp(id);
    }
}

void Remove(int &id, int v, int num) {
    if(!id)
        return;
    else {
        if(v == val[id]) {
            if(cnt[id] > num) {
                cnt[id -= num];
                PushUp(id);
            } else if(ch[id][0] || ch[id][1]) {
                if(!ch[id][1] || dat[ch[id][0]] > dat[ch[id][1]])
                    Rotate(id, 1), Remove(ch[id][1], v, num);
                else
                    Rotate(id, 0), Remove(ch[id][0], v, num);
                PushUp(id);
            } else {
                st[++stop] = id;
                id = 0;
            }
        } else {
            v < val[id] ? Remove(ch[id][0], v, num) : Remove(ch[id][1], v, num);
            PushUp(id);
        }
    }
}

int GetRank(int id, int v) {
    //小于等于v的有几个？
    if(!id)
        return 0;
    else {
        if(v == val[id])
            return siz[ch[id][0]] + cnt[id];
        else if(v < val[id])
            return GetRank(ch[id][0], v);
        else
            return siz[ch[id][0]] + cnt[id] + GetRank(ch[id][1], v);
    }
}

//把id2的树整棵插进id1中
void Merge(int &id1, int &id2) {
    if(siz[id1] < siz[id2])
        swap(id1, id2);
    //将id2的根插进id1，目前每次只搬运一个
    while(siz[id2]) {
        int tmpv = val[id2], tmpn = cnt[id2];
        Remove(id2, tmpv, tmpn);
        Insert(id1, tmpv, tmpn);
    }
    return;
}

int d[MAXN], f[MAXN], a[MAXN];

int u2topo[MAXN], cnttopo, topo2u[MAXN];

struct Query {
    int u, k, ans, id;
    bool operator<(const Query& q)const {
        return u2topo[u] < u2topo[q.u];
    }
} que[MAXN];

struct cmp {
    bool operator()(const Query& q1, const Query& q2)const {
        return q1.id < q2.id;
    }
};

int que2[MAXN];
int front, back;

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, q;
    while(~scanf("%d%d", &n, &q)) {
        memset(d, 0, sizeof(d[0]) * (n + 1));
        for(int i = 2; i <= n; ++i) {
            scanf("%d", &f[i]);
            ++d[f[i]];
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        front = 1, back = 0, cnttopo = 0;
        for(int i = 1; i <= n; ++i) {
            if(d[i] == 0)
                que2[++back] = i;
        }
        while(back >= front) {
            int u = que2[front++];
            u2topo[u] = ++cnttopo;
            topo2u[cnttopo] = u;
            d[f[u]]--;
            if(d[f[u]] == 0)
                que2[++back] = f[u];
        }
        for(int i = 1; i <= q; ++i) {
            scanf("%d%d", &que[i].u, &que[i].k);
            que[i].id = i;
        }
        sort(que + 1, que + 1 + q);
        Init(n);
        for(int i = 1; i <= n; ++i) {
            Insert(root[i], a[i], 1);
            //cout<<siz[root[i]]<<endl;
            //cout<<u2topo[i]<<" ";
        }
        //cout<<endl;
        int curtopo = 1;
        for(int i = 1; i <= q; ++i) {
            while(u2topo[que[i].u] > curtopo) {
                //cout<<"Qu="<<que[i].u<<" topoQu="<<u2topo[que[i].u]<<endl;
                //等于的话不用合并，直接查询
                int u = topo2u[curtopo++];
                //cout<<"Merge "<<u<<" "<<f[u]<<endl;
                Merge(root[f[u]], root[u]);
                //cout<<"  After Merge Size="<<siz[root[f[u]]]<<endl;
            }
            //现在相等了
            que[i].ans = GetRank(root[que[i].u], que[i].k);
            /*cout<<"Answer u="<<que[i].u;
            cout<<" ans="<<que[i].ans<<endl;*/
        }
        sort(que + 1, que + 1 + q, cmp());
        for(int i = 1; i <= q; ++i) {
            printf("%d
", que[i].ans);
        }
        //puts("---");
    }
    return 0;
}