Tree and Queries 【CodeForces - 375D】【Dsu on Tree + 树状数组】

题目链接

---

  现有一棵 n 个点的树，点的编号从 1 起，树以 1 为根，每个点 i 都一个颜色 ci，接下来有 m 个询问，每次询问以 vj 为根的子树中，求有多少种颜色，这些颜色在子树中出现的次数至少为 kj。

  首先，想到是维护子树，我们就往Dsu上面靠，再者，又是要维护颜色个数大于等于K的，所以我们就要想办法把颜色维护，我的做法是，我们可以确定放进子树中的颜色的种类数，然后呢，我们现在要得到出现次数至少为k的，所以我们可以维护用总的颜色数减去出现次数小于等于K-1的颜色的种类数。这里就是通过树状数组维护了。

  所以，总的复杂度就是。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, Q, head[maxN], cnt, col[maxN];
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
struct Question
{
    int kth, ith;
    Question(int a=0, int b=0):kth(a), ith(b) {}
    friend bool operator < (Question e1, Question e2) { return e1.kth < e2.kth; }
};
vector<Question> vt[maxN];
int siz[maxN], dfn[maxN], tot, rid[maxN], Wson[maxN];
void pre_dfs(int u, int fa)
{
    siz[u] = 1; dfn[u] = ++tot; rid[tot] = u;
    int maxx = 0;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa) continue;
        pre_dfs(v, u);
        siz[u] += siz[v];
        if(siz[v] > maxx)
        {
            maxx = siz[v];
            Wson[u] = v;
        }
    }
}
int ans[maxN] = {0}, col_siz[maxN] = {0}, num[maxN] = {0}, col_num, trie[maxN] = {0};
inline void update(int x, int val) { while(x <= 100000) { trie[x] += val; x += lowbit(x); } }
inline int query(int x) { int sum = 0; while(x) { sum += trie[x]; x -= lowbit(x); } return sum; }
void dfs(int u, int fa, bool keep)
{
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa || v == Wson[u]) continue;
        dfs(v, u, false);
    }
    if(Wson[u])
    {
        dfs(Wson[u], u, true);
    }
    for(int i=head[u], v, id; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa || v == Wson[u]) continue;
        for(int j=dfn[v]; j<dfn[v] + siz[v]; j++)
        {
            id = rid[j];
            if(col_siz[col[id]]) update(col_siz[col[id]], -1);
            else col_num++;
            col_siz[col[id]]++;
            update(col_siz[col[id]], 1);
        }
    }
    if(col_siz[col[u]]) update(col_siz[col[u]], -1);
    else col_num++;
    col_siz[col[u]]++;
    update(col_siz[col[u]], 1);
    int len = (int)vt[u].size();
    Question now;
    for(int i=0; i<len; i++)
    {
        now = vt[u][i];
        ans[now.ith] = col_num - query(now.kth - 1);
    }
    if(!keep)
    {
        for(int i=dfn[u], id; i<dfn[u] + siz[u]; i++)
        {
            id = rid[i];
            update(col_siz[col[id]], -1);
            col_siz[col[id]]--;
            if(col_siz[col[id]]) update(col_siz[col[id]], 1);
            else col_num--;
        }
    }
}
inline void init()
{
    cnt = tot = col_num = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &Q);
    init();
    for(int i=1; i<=N; i++) scanf("%d", &col[i]);
    for(int i=1, u, v; i<N; i++)
    {
        scanf("%d%d", &u, &v);
        _add(u, v);
    }
    pre_dfs(1, 0);
    for(int i=1, vi, ki; i<=Q; i++)
    {
        scanf("%d%d", &vi, &ki);
        vt[vi].push_back(Question(ki, i));
    }
    dfs(1, 0, false);
    for(int i=1; i<=Q; i++) printf("%d\n", ans[i]);
    return 0;
}