dfs序题集

dfs序可以维护一个子树内的信息

需要记录dfs进的时间以及所有子树都遍历完的时间

void dfs(int u, int fa)
{
    L[u] = ++id;
    for(int i = head[u]; ~i; i = e[i].next)
    {
        int v = e[i].to;
        if(v == fa) continue;
        dfs(v, u);
    }
    R[u] = id;
}

那么对于点i，L[i]到R[i]就是i的子树（包括i）

那么子树内维护信息就可以用树状数组，线段树之内的乱搞了。

poj 3321

用树状数组维护就好。

注意修改的时候一定是修改序列

#include<cstdio>
#include<cstring>
#include<cctype>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e5 + ;
struct Edge{ int to, next; };
Edge e[MAXN << ];
int head[MAXN], num, n, m;
int L[MAXN], R[MAXN], s[MAXN], id;

struct Binary_Index_Tree
{
    int f[MAXN];

    Binary_Index_Tree() { memset(f, , sizeof(f)); }

    int lowbit(int x) { return x & (-x); }

    void add(int x, int p)
    {
        while(x <= n)
        {
            f[x] += p;
            x += lowbit(x);
        }
    }

    int sum(int x)
    {
        int res = ;
        while(x)
        {
            res += f[x];
            x -= lowbit(x);
        }
        return res;
    }

    int query(int l, int r) { return sum(r) - sum(l - ); }
}S;

void read(int& x)
{
    int f = ; x = ; char ch = getchar();
    while(!isdigit(ch)) { if(ch == '-') f = -; ch = getchar(); }
    while(isdigit(ch)) { x = x *  + ch - ''; ch = getchar(); }
    x *= f;
}

void AddEdge(int to, int from)
{
    e[num] = Edge{to, head[from]};
    head[from] = num++;
}

void dfs(int u, int fa)
{
    L[u] = ++id;
    for(int i = head[u]; ~i; i = e[i].next)
    {
        int v = e[i].to;
        if(v == fa) continue;
        dfs(v, u);
    }
    R[u] = id;
}

int main()
{
    memset(head, -, sizeof(head)); num = ;
    read(n);
    REP(i, , n)
    {
        int u, v; read(u), read(v);
        AddEdge(u, v); AddEdge(v, u);
    }
    _for(i, , n)
    {
        s[i] = ;
        S.add(i, );
    }

    dfs(, -);
    read(m);

    while(m--)
    {
        char op[]; int x;
        scanf("%s", op); read(x);
        if(op[] == 'Q') printf("%d\n", S.query(L[x], R[x]));
        else
        {
            int id = L[x];
            if(s[id]) S.add(id, -);
            else S.add(id, );
            s[id] = !s[id];
        }
    }

    return ;
}