小A的礼物【Dsu on Tree】

题目链接

  求子树信息，这样一类的题目，往往可以使用Dsu on Tree这样的启发式合并的做法来实现。

  我们直接对于轻儿子每次均不保留，并且先查轻儿子，再查重儿子，以此来满足时间复杂度。

#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 eps 1e-4
#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
#define MP(a, b) make_pair(a, b)
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];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
int siz[maxN], Wson[maxN] = {0};
void pre_dfs(int u)
{
siz[u] = 1;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
pre_dfs(v);
if(siz[v] > siz[Wson[u]]) Wson[u] = v;
siz[u] += siz[v];
}
}
int num[maxN] = {0}, have = 0, ans[maxN], dfn[maxN], rid[maxN], tot = 0;
void dfs(int u, bool keep)
{
dfn[u] = ++tot; rid[tot] = u;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == Wson[u]) continue;
dfs(v, false);
}
if(Wson[u])
{
dfs(Wson[u], true);
}
if(!num[col[u]]) have++;
num[col[u]]++;
for(int i=head[u], v, id; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == Wson[u]) continue;
for(int j=dfn[v]; j<dfn[v] + siz[v]; j++)
{
id = rid[j];
if(!num[col[id]]) have++;
num[col[id]] ++;
}
}
ans[u] = have;
if(!keep)
{
for(int i=dfn[u], id; i<dfn[u] + siz[u]; i++)
{
id = rid[i];
num[col[id]]--;
if(!num[col[id]]) have--;
}
}
}
inline void init()
{
cnt = 0;
for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
scanf("%d", &N); init();
for(int i=2, ff; i<=N; i++)
{
scanf("%d", &ff);
addEddge(ff, i);
}
for(int i=1; i<=N; i++) scanf("%d", &col[i]);
pre_dfs(1);
dfs(1, true);
scanf("%d", &Q); int op;
while(Q--)
{
scanf("%d", &op);
printf("%dn", ans[op]);
}
return 0;
}

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