HDU 3887 Counting Offspring (树状数组+人工模拟栈)

对这棵树DFS遍历一遍，同一节点入栈和出栈之间访问的节点就是这个节点的子树。

因此节点入栈时求一次 小于 i 的节点个数 和，出栈时求一次 小于 i 的节点个数 和，两次之差就是答案。

PS.这题直接DFS会爆栈，可以重新设置栈的大小

#pragma comment(linker,"/STACK:100000000,100000000")

也可以人工模拟栈，代码如下。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 110000;

struct Edge
{
    int v;
    int next;
};

int N, root;
int EdgeN, top;
Edge D[ (MAXN << 2) + 100 ];
int C[ MAXN ];     //树状数组
int ans[ MAXN ];   //答案
int pre[ MAXN ];   //入点前的和
int head[ MAXN ];
int stack[MAXN];   //手工栈
bool vis[MAXN];

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

int query( int x )
{
    int res = 0;
    while ( x > 0 )
    {
        res += C[x];
        x -= lowbit(x);
    }
    return res;
}

void update( int x, int val )
{
    while ( x <= N )
    {
        C[x] += val;
        x += lowbit(x);
    }
    return;
}

void AddEdge( int u, int v )
{
    D[EdgeN].v = v;
    D[EdgeN].next = head[u];
    head[u] = EdgeN++;
    return;
}

void showStack( int top )
{
    for ( int i = 1; i <= top; ++i )
        printf( "%d ", stack[i] );
    puts("
===================");
    return;
}

void DFS( int u )
{
    stack[++top] = u;
    vis[u] = true;
    while ( top )
    {
        int uu = stack[top];
        if ( !vis[uu] ) pre[uu] = query(uu - 1);
        vis[uu] = true;
        int i;
        for ( i = head[uu]; i != -1; i = D[i].next )
        {
            if ( !vis[ D[i].v ] )
            {
                update( D[i].v, 1 );
                stack[++top] = D[i].v;
                //showStack(top);
                break;
            }
        }
        if ( i == -1 )
        {
            ans[uu] = query( uu - 1 ) - pre[uu];
            --top;
        }
    }
    return;
}

int main()
{
    //freopen( "s.txt", "w", stdout );
    while ( scanf( "%d%d", &N, &root ), N || root )
    {
        EdgeN = 1;
        memset( head, -1, sizeof(head) );
        memset( C, 0, sizeof(C) );

        for ( int i = 1; i < N; ++i )
        {
            int u, v;
            scanf( "%d%d", &u, &v );
            AddEdge( u, v );
            AddEdge( v, u );
        }

        top = 0;
        memset( vis, false, sizeof(vis) );
        memset( pre, 0, sizeof(pre) );
        DFS( root );

        printf( "%d", ans[1] );
        for ( int i = 2; i <= N; ++i )
            printf( " %d", ans[i] );
        puts("");
    }
    return 0;
}