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): 3284    Accepted Submission(s): 1118

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 77 107 17 97 37 410 1414 214 139 119 66 56 83 153 120 0

 

Sample Output

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

 

Author

bnugong

 

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<map>#include<vector>#include<queue>#include<stack>#define eps 1e-8const int inf = 0x3f3f3f3f;const long long mod=1e9+7;const int N=100020;using namespace std;int n,p,cnt,k;struct node{    int v,nxt;}s[N*2];int first[N];int in[N],out[N];int ans[N];int ar[N];int lowbit(int x){    return x&-x;}void add(int i,int w){    while(i<=n)    {        ar[i]+=w;        i+=lowbit(i);    }}int sum(int k){    int ans=0;    while(k)    {        ans+=ar[k];        k-=lowbit(k);    }    return ans;}int dfs(int u,int pre){        in[u]=++cnt;    for(int i=first[u];i;i=s[i].nxt)    {        int v=s[i].v;        if(v==pre)            continue;        dfs(v,u);    }    out[u]=cnt;}int main(){    int u,c;    while(~scanf("%d%d",&n,&p)&&(n+p))    {        memset(first,0,sizeof(first));        memset(ar,0,sizeof(ar));        k=1,cnt=0;        for(int i=1;i<n;i++)        {           scanf("%d%d",&u,&c);        s[k].v = c;        s[k].nxt = first[u];        first[u] = k++;        s[k].v=u;        s[k].nxt=first[c];        first[c]=k++;        }        dfs(p,-1);       for(int i=1;i<=n;i++)       {           int rr=sum(out[i])-sum(in[i]-1);           printf("%d%c",rr,((i==n)?'\n':' '));           add(in[i],1);       }    }}