HDU 3887 Counting Offspring（DFS序+树状数组）

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3054    Accepted Submission(s): 1031

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：求在一个有根节点的树中 每个节点的子树中有多少个小于改点的序号

DFS序找到每个节点的时间戳  用树状数组直接维护这个区间就可以了

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cctype>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 #include<algorithm>
 #include<string>
 #define ll long long
 #define eps 1e-10
 #define LL unsigned long long
 using namespace std;
 const int inf=0x3f3f3f3f;
 const int N=+;
 const int mod=1e9+;
 int head[N];
 int tot,time;
 int L[N],R[N];
 struct node{
     int to,next;
 }edge[N<<];
 int a[N*];
 void init(){
     memset(head,-,sizeof(head));
     tot=;
     time=;
 }
 void add(int u,int v){
     edge[tot].to=v;
     edge[tot].next=head[u];
     head[u]=tot++;
 }
 void DFS(int x,int fa){
     L[x]=++time;
     for(int i=head[x];i!=-;i=edge[i].next){
         int v=edge[i].to;
         if(v==fa)continue;
         DFS(v,x);
     }
     R[x]=time;

 }
 int lowbit(int x){
     return x&(-x);
 }
 void update(int x,int y){
     while(x<=time){
         a[x]=a[x]+y;
         x=x+lowbit(x);
     }
 }
 int getsum(int x){
     int ans=;
     while(x>){
         //cout<<2<<endl;
         ans=ans+a[x];
         x=x-lowbit(x);
     }
     return ans;
 }
 int main(){
     int n,p;
     while(scanf("%d%d",&n,&p)!=EOF){
         if(n==&&p==)break;
         init();
         memset(a,,sizeof(a));
         for(int i=;i<n;i++){
             int u,v;
             scanf("%d%d",&u,&v);
             add(u,v);
             add(v,u);
         }
         DFS(p,-);
         for(int i=;i<=n;i++){
             int ans=getsum(R[i])-getsum(L[i]-);
             if(i!=n)cout<<ans<<" ";
             else{
                 cout<<ans<<endl;
             }
             update(L[i],);
         }
     }
 }