hdu3887 Counting Offspring

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1148    Accepted Submission(s): 406

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

 

Source

2011 Multi-University Training Contest 5 - Host by BNU

 

Recommend

lcy

 

照样是带时间戳的树状数组，记得去年多校的时候宣姐死活YY不出来……orz，然后我提供了思路，但是不会写，今天终于把他写完了。

注意直接深搜会爆栈，于是乎学了下栈模拟递归的写法……orz。

代码

#include <stdio.h>#include <string.h>#include <stack>#include <vector>#include <algorithm>using namespace std;stack <int> s;int vis[100005];int in[100005];int ans[100005];int n;vector <int> v[100005];int lowbit(int t){    return t&(-t);}int Query(int t){    int sum=0;    while(t>0)    {        sum+=in[t];        t-=lowbit(t);    }    return sum;}void Add(int t,int val){    while(t<=n)    {        in[t]+=val;        t+=lowbit(t);    }}void DFS(int t){    int tag;    s.push(t);    while(!s.empty())    {        tag=s.top();        if (!vis[tag])        {            vis[tag]=1;            ans[tag]=Query(tag);        }        else if (v[tag].size()!=0)        {            if (vis[v[tag].back()]==0) s.push(v[tag].back());            v[tag].pop_back();        }        else        {            ans[tag]=Query(tag)-ans[tag];            s.pop();            Add(tag,1);        }    }}int main(){    int m,i,j,T,x,y;    while(1)    {        scanf("%d%d",&n,&m);        if (n==0 && m==0) break;        for (i=0;i<=n;i++)        {            v[i].clear();        }        for (i=0;i<n-1;i++)        {            scanf("%d%d",&x,&y);            v[x].push_back(y);            v[y].push_back(x);        }        memset(in,0,sizeof(in));        memset(vis,0,sizeof(vis));        memset(ans,0,sizeof(ans));        DFS(m);        for (i=1;i<n;i++)        {            printf("%d ",ans[i]);        }        printf("%d\n",ans[i]);    }    return 0;}