HDU 3887 Counting Offspring ［DFS序列＋树状数组］

HDU 3887 Counting Offspring ［DFS序列＋树状数组］

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 423    Accepted Submission(s): 137

Problem Description

You are given a tree, it’s root is p, and the node is numbered from 1 to n. Now define f(i) as the number of nodes whose number is less than i in all the succeeding nodes of node i. Now we need to calculate f(i) for any possible i.

 

Input

Multiple cases (no more than 10), for each case:
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros.

 

Output

For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.

 

Sample Input

    
    
    
    

     
     
     
     15 7
7 10
7 1
7 9
7 3
7 4
10 14
14 2
14 13
9 11
9 6
6 5
6 8
3 15
3 12
0 0
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     0 0 0 0 0 1 6 0 3 1 0 0 0 2 0
    
    
    
    

 

Author

bnugong

 

Source

2011 Multi-University Training Contest 5 - Host by BNU

题意是，给一棵树，给每个结点一个标号，问所有存在的结点，它的子树中，结点标号小于它的有多少个。

做法：DFS序列构造＋树状数组查询区间。

竟然忘了DFS的遍历的意义。对一棵树进行DFS遍历，一个结点最多到达两次，一次出，一次入，而这之间的结点标号均为它的子树结点的标号。所以只要用某种方法快速查询其左右标号内含的区间中，比它小的标号个数即可。（别＊2）在肖神的提携下，ym到了树状数组在本题的使用方法。

做法流程。初始化（记得树状数组和DFS记录序列要开点数*2的。。因为这个nc半天，不解释。） －>建图－> DFS（因为HDU坑爹的栈大小，必须手写栈模拟DFS才能过。。第一次手工模拟DFS。。太蛋疼了。。调了无数次。）－>对于每个点查询DFS序列。（既然是离线查询，那么就按照从小到大查吧。凡事要有个序，有序了就好查了。对于每个点的标号，先查它的左右区间是否有比它小的sum(r[i] - 1) - sum(l[i]]).....再把它插入树状数组中，插一次就好，不然结果还得除二。（都是肖神！）
T_T。。记得树状数组和dfs序列都是开到点数*2的，而其他的都是点数大小的。注意初始化和树状数组查询和更新时的范围同理。
代码：

#include  < iostream >
#include  < cstdio >
#include  < cstring >
#include  < stack >
#include  < algorithm >
using   namespace  std;
stack  < int >  S;
#define  maxn 100005
int  cnt,p,n,tot,lit[maxn],seq[maxn  *   2 ],l[maxn],r[maxn],tr[maxn  *   2 ];
int  bit( int  x){ return  x  &   - x;}
bool  vis[maxn];
struct  edge
{
     int  p,next;
    edge(){}
    edge( int  a, int  b):p(a),next(b){}
}v[maxn],e[maxn  *   2 ];
void  add( int  p, int  q)
{
    e[tot]  =  edge(q,v[p].next);
    v[p].next  =  tot ++ ;
}
int  sum( int  x)
{
     int  ans  =   0 ;
     for (;x  >   0 ;x  -=  bit(x))
        ans  +=  tr[x];
     return  ans;
}
void  mod( int  x, int  val)
{
     for (;x  <  cnt;x  +=  bit(x))
        tr[x]  +=  val;    
}
void  init()
{
    tot  =   0 ;
     for ( int  i  =   1 ;i  <=  n;i ++ )
        v[i].next  =   - 1 ;
    fill(tr,tr  +   2   *  n  +   1 , 0 );
    fill(vis,vis  +  n  +   1 , 0 );
     while ( ! S.empty())
        S.pop();
}
void  dfs()
{
    cnt  =   1 ;
    S.push(p);
     while ( ! S.empty())
    {
         int  now  =  S.top();
         if ( ! vis[now])
        {
            seq[cnt]  =  now;
            l[now]  =  cnt ++ ;
            vis[now]  =   true ;
        }
         bool  mark  =   false ;
         for ( int  k  =  v[now].next; ~ k;k  =  e[k].next)
        {
             if ( ! vis[e[k].p])
            {
                mark  =   true ;
                S.push(e[k].p);
                v[now].next  =  e[k].next;
                 break ;
            }
        }
         if (mark)
             continue ;
         if (vis[now])
        {
            seq[cnt]  =  now;
            r[now]  =  cnt ++ ;
            S.pop();
        }
    }
}
void  gao()
{
     for ( int  i  =   1 ;i  <=  n;i ++ )
    {
        lit[i]  =  sum(r[i]  -   1 )  -  sum(l[i]);
        mod(l[i], 1 );
    }
}
int  main()
{
     while (scanf( " %d %d " , & n, & p)  ==   2   &&  (p  ||  n))
    {
        init();
         for ( int  i  =   0 ;i  <  n  -   1 ;i ++ )
        {
             int  a,b;
            scanf( " %d %d " , & a, & b);
            add(a,b);
            add(b,a);
        }
        dfs();
        gao();
         for ( int  i  =   1 ;i  <=  n;i ++ )
        {
            printf( " %d " ,lit[i]);
            putchar((i  ==  n) ? ' \n ' : '   ' );
        }
    }
}

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