cf246E. Blood Cousins Return

Posted 2021-09-17 Jozky86

cf246E. Blood Cousins Return

题意：

给你一个森林，每个点都有自己的种类，问以v为根节点的子树中，与v距离为k的节点有多少种

题解：

和cf208E. Blood Cousins这个题差不多，就是多了一个种类，用一个unordered_map对名字进行编号，用map对每一层的名字进行标记，(能用unordered_map的就不要用map，不然后超时)
详细看代码

代码：

// Problem: E. Blood Cousins Return
// Contest: Codeforces - Codeforces Round #151 (Div. 2)
// URL: https://codeforces.com/contest/246/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// Data:2021-09-02 17:37:18
// By Jozky

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
    startTime= clock();
    freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 1e5 + 9;
int n, m;
vector<int> vec[maxn];
vector<PII> q[maxn];
unordered_map<string, int> mp;
unordered_map<int, string> na;
int f[maxn][30];
int son[maxn];
int Son;
int dep[maxn], siz[maxn];
void dfs1(int u, int fa)
{
    dep[u]= dep[fa] + 1;
    siz[u]= 1;
    f[u][0]= fa;
    for (int i= 1; i <= 20; i++)
        f[u][i]= f[f[u][i - 1]][i - 1];
    for (auto v : vec[u]) {
        if (v == fa)
            continue;
        dfs1(v, u);
        siz[u]+= siz[v];
        if (siz[v] > siz[son[u]])
            son[u]= v;
    }
}
int find_f(int u, int k)
{
    for (int i= 0; i <= 20; i++) {
        if ((1 << i) & k)
            u= f[u][i];
    }
    return u;
}
map<pair<int, int>, int> iff;
// int iff[maxn][200];
int ans[maxn];
int num[maxn];
void add(int u, int fa, int val)
{
    int id= mp[na[u]];
    //    cout<<"name="<<na[u]<<" id="<<id<<endl;
    if (val == 1) {
        iff[{id, dep[u]}]++;
        if (iff[{id, dep[u]}] == 1)
            num[dep[u]]+= val;
    }
    else if (val == -1) {
        iff[{id, dep[u]}]--;
        if (iff[{id, dep[u]}] == 0)
            num[dep[u]]+= val;
    }
    for (auto v : vec[u]) {
        if (v == fa || v == Son)
            continue;
        add(v, u, val);
    }
}
void dfs2(int u, int fa, int keep)
{
    for (auto v : vec[u]) {
        if (v == fa || v == son[u])
            continue;
        dfs2(v, u, 0);
    }
    if (son[u]) {
        dfs2(son[u], u, 1);
        Son= son[u];
    }
    add(u, fa, 1);
    for (auto it : q[u]) {
        int deep= it.first + dep[u];
        int id= it.second;
        ans[id]= max(0, num[deep]);
    }
    Son= 0;
    if (!keep) {
        add(u, fa, -1);
    }
}
int main()
{
    //rd_test();
    read(n);
    for (int i= 1; i <= n; i++) {
        string name;
        int x;
        cin >> name >> x;
        // if(mp[name]!=0)
        na[i]= name;
        mp[name]= i;
        vec[x].push_back(i);
    }
    dfs1(0, 0);
    read(m);
    for (int i= 1; i <= m; i++) {
        int v, k;
        read(v, k);
        //        int f= find_f(v, k);
        q[v].push_back({k, i});
    }
    dfs2(0, 0, 0);
    for (int i= 1; i <= m; i++)
        printf("%d\\n", ans[i]);
    //Time_test();
}