zoj 3686 A Simple Tree Problem （线段树）

Solution：

　　根据树的遍历道的时间给树的节点编号，记录下进入节点和退出节点的时间。这个时间区间覆盖了这个节点的所有子树，可以当做连续的区间利用线段树进行操作。

/*
       线段树
*/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#define lson x<<1
#define rson x<< 1 | 1
using namespace std;

const int MAX = ;
int w[MAX], to[MAX];
int tl[MAX << ], tr[MAX << ], flag[MAX << ], sum[MAX << ][];
int l, r;
struct Edge {
    int v, ne;
} edge[MAX << ];

int head[MAX], cnt, num;

void Dfs ( int u )
{
    w[u] = ++num;
    for ( int i = head[u]; i != ; i = edge[i].ne ) {
        Dfs ( edge[i].v );
    }
    to[u] = num;
}

void addE ( int u, int v )
{
    edge[++cnt].v = v;
    edge[cnt].ne = head[u], head[u] = cnt;
}

void build ( int x, int l, int r )
{
    flag[x] = sum[x][] = sum[x][] = ;
    tl[x] = l, tr[x] = r;
    if ( l == r ) {
        sum[x][] = ;
        return ;
    }
    int mid = ( l + r ) >> ;
    build ( lson, l, mid ), build ( rson, mid + , r );
    sum[x][] += sum[lson][] + sum[rson][];
}

void push ( int x )
{
    if ( flag[x] ==  ) return;
    flag[x] = ;
    flag[lson] ^= ;
    swap ( sum[lson][], sum[lson][] );
    flag[rson] ^= ;
    swap ( sum[rson][], sum[rson][] );
}
void updata ( int x )
{
    sum[x][] = sum[lson][] + sum[rson][];
    sum[x][] = sum[lson][] + sum[rson][];
}
void modify ( int x )
{
    if ( tl[x] >= l && tr[x] <= r ) {
        flag[x] ^= ;
        swap ( sum[x][], sum[x][] );
        return ;
    }
    push ( x );
    int mid = ( tl[x] + tr[x] ) >> ;
    if ( l <= mid ) modify ( lson );
    if ( r > mid ) modify ( rson );
    updata ( x );
}

int query ( int x )
{
    if ( tl[x] >= l && tr[x] <= r ) {
        return sum[x][];
    }
    push ( x );
    int mid = ( tl[x] + tr[x] ) >> , ans = ;
    if ( l <= mid ) ans += query ( lson );
    if ( r > mid ) ans += query ( rson );
    updata ( x );
    return ans;
}

int n, m;

void init()
{
    num = cnt = ;
    memset ( head, , sizeof head );
}
int main()
{
    while ( scanf ( "%d %d", &n, &m ) != EOF ) {
        init();
        for ( int i = , v; i <= n; i++ ) {
            scanf ( "%d", &v );
            addE ( v, i );
        }
        Dfs (  );
        scanf ( "%d", &m );
        build ( , , n );
        char cmd[];
        for ( int i = , u; i <= m; ++i ) {
            scanf ( "%s %d", cmd, &u );
            l = w[u], r = to[u];
            if ( cmd[] == 'o' ) {
                modify (  );
            } else {
                printf ( "%d\n", query (  ) );
            }
        }
        putchar();
    }
}