HDU 3887 Counting Offspring

Counting Offspring

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

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

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#define pb push_back
using namespace std;
const int N=1e5+5;
int n,rt,dfs_clock,L
,R
,sum
,ans
;
vector<int>G
;
inline void init()
{
memset(sum,0,sizeof(sum));
for(int i=1;i<=n;i++)G[i].clear();
dfs_clock=0;
}
void dfs(int u,int fa)
{
L[u]=++dfs_clock;
for(int len=G[u].size(),i=0;i<len;i++)
{
int v=G[u][i];
if(v==fa)continue;
dfs(v,u);
}
R[u]=dfs_clock;
}
inline void add(int pos)
{
for(int i=pos;i<=n;i+=i&(-i))
sum[i]+=1;
}
inline int query(int pos)
{
int res=0;
for(int i=pos;i>0;i-=i&(-i))
res+=sum[i];
return res;
}
int main()
{
while(scanf("%d%d",&n,&rt),n|rt)
{
init();
for(int _=0;_<n-1;_++)
{
int u,v;
scanf("%d%d",&u,&v);
G[u].pb(v),G[v].pb(u);
}
dfs(rt,-1);
for(int i=1;i<=n;i++)
{
ans[i]=query(R[i])-query(L[i]-1);
add(L[i]);
}
for(int i=1;i<=n;i++)
printf(i==n?"%d\n":"%d ",ans[i]);
}
return 0;
}