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=200000+10;
const int mod=1e9+7;
int head[N];
int tot,time;
int L[N],R[N];
struct node{
    int to,next;
}edge[N<<1];
int a[N*5];
void init(){
    memset(head,-1,sizeof(head));
    tot=0;
    time=0;
}
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!=-1;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=0;
    while(x>0){
        //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==0&&p==0)break;
        init();
        memset(a,0,sizeof(a));
        for(int i=1;i<n;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        DFS(p,-1);
        for(int i=1;i<=n;i++){
            int ans=getsum(R[i])-getsum(L[i]-1);
            if(i!=n)cout<<ans<<" ";
            else{
                cout<<ans<<endl;
            }
            update(L[i],1);
        }
    }
}