[Codeforces 877E] Danil and a Part-time Job

[题目链接]

        https://codeforces.com/contest/877/problem/E

[算法]

        首先求出这棵树的DFS序

        一棵子树的DFS序为连续的一段 ， 根据这个性质 ， 用线段树维护区间中1的个数即可

        时间复杂度 : O(NlogN)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
typedef long long ll;
typedef long double ld;

struct edge
{
        int to , nxt;
} e[MAXN << 2];

int n , q , timer , tot;
int val[MAXN] , l[MAXN] , r[MAXN] , dfn[MAXN] , size[MAXN] , head[MAXN] , loc[MAXN] , par[MAXN];

struct Segment_Tree
{
        struct Node
        {
                int l , r , cnt;
                bool tag;        
        }    a[MAXN << 2];
        inline void update(int index)
        {
                a[index].cnt = a[index << 1].cnt + a[index << 1 | 1].cnt;
        }
        inline void build(int index , int l , int r)
        {
                a[index].l = l , a[index].r = r;
                a[index].tag = false;
                if (l == r)
                {
                        a[index].cnt = (val[loc[l]] == 1);
                        return;
                }
                int mid = (l + r) >> 1;
                build(index << 1 , l , mid);
                build(index << 1 | 1 , mid + 1 , r);
                update(index);
        }
        inline void pushdown(int index)
        {
                int l = a[index].l , r = a[index].r;
                int mid = (l + r) >> 1;
                if (a[index].tag)
                {
                        a[index << 1].cnt = (mid - l + 1) - a[index << 1].cnt;
                        a[index << 1 | 1].cnt = (r - mid) - a[index << 1 | 1].cnt;
                        a[index << 1].tag ^= 1; 
                        a[index << 1 | 1].tag ^= 1;
                        a[index].tag = false; 
                }
        }
        inline void modify(int index , int l , int r)
        {
                if (a[index].l == l && a[index].r == r)
                {
                        a[index].cnt = (r - l + 1) - a[index].cnt;
                        a[index].tag ^= 1;
                        return;
                }
                pushdown(index);
                int mid = (a[index].l + a[index].r) >> 1;
                if (mid >= r) modify(index << 1 , l , r);
                else if (mid + 1 <= l) modify(index << 1 | 1 , l , r);
                else
                {
                        modify(index << 1 , l , mid);
                        modify(index << 1 | 1 , mid + 1 , r);
                }
                update(index);
        }
        inline int query(int index , int l , int r)
        {
                if (a[index].l == l && a[index].r == r)
                        return a[index].cnt;
                pushdown(index);
                int mid = (a[index].l + a[index].r) >> 1;
                if (mid >= r) return query(index << 1 , l , r);
                else if (mid + 1 <= l) return query(index << 1 | 1 , l , r);
                else return query(index << 1 , l , mid) + query(index << 1 | 1 , mid + 1 , r);
        }
} SGT;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u , int v)
{
        ++tot;
        e[tot] = (edge){v , head[u]};
        head[u] = tot;
}
inline void dfs(int u)
{
        l[u] = dfn[u] = ++timer;
        loc[timer] = u;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (v == par[u]) continue;
                dfs(v);                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
        }        
        r[u] = timer;
}

int main()
{
        
        scanf("%d" , &n);
        for (int i = 2; i <= n; i++)
        {
                scanf("%d" , &par[i]);
                addedge(par[i] , i);        
        }
        for (int i = 1; i <= n; i++) scanf("%d" , &val[i]);
        dfs(1);
        SGT.build(1 , 1 , n);
        scanf("%d" , &q);
        while (q--)
        {
                char type[5];
                scanf("%s" , type);
                if (type[0] == 'p')
                {
                        int u;
                        scanf("%d" , &u);
                        SGT.modify(1 , l[u] , r[u]);
                } else
                {
                        int u;
                        scanf("%d" , &u);
                        printf("%d
" , SGT.query(1 , l[u] , r[u]));
                }
        }
        
        return 0;
    
}