HDU

Counting Offspring

 HDU - 3887

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

Source

HDU - 3887

My Solution

题意：问对于每个节点，它的子树上标号比它小的点有多少个。

dfs序+线段树

关于dfs序：

dfs序是处理树上问题很重要的一个工具，主要能够解决对于一个点，它的子树上的一些信息的维护，即用来处理子树的问题。 dfs序一般开的空间是n，因为只在入的地方时间戳++，出来的地方时间戳没有额外的++，线段树的每个节点应当是时间戳。

这里，因为点在它的子树上，所以在线段树中，故在它的两个时间戳的区间内([p1[i],p2[i])，所以我们只需要从小到大考虑，它的区间里有多少个点已经放入，然后再把它放入即可。

时间复杂度 O(nlogn)

空间复杂度 O(4*n)

#include <iostream>#include <cstdio>#include <vector>#include <cstring>using namespace std;typedef long long LL;const int MAXN = 1e5 + 8;vector<int> sons[MAXN];//dfs序int p1[MAXN], p2[MAXN], ti = 0;int dfsnum[MAXN];  //这个按情况是否需要。inline void get_dfs_list(int u, int fa){    p1[u] = ++ti;    dfsnum[ti] = u; //    int sz = sons[u].size(), i, v;    for(i = 0; i < sz; i++){        v = sons[u][i];        if(v == fa) continue;        get_dfs_list(v, u);    }    p2[u] = ti;}//线段树int sum[4*MAXN];int size;inline void pushup(int Ind){    sum[Ind] = sum[Ind<<1] + sum[(Ind<<1) + 1];}inline int _Query(int a, int b, int l, int r, int Ind){    if(a <= l && b >= r) return sum[Ind];    int mid = (l+r)>>1; int ret = 0;    if(a <= mid) ret += _Query(a, b, l, mid, Ind<<1);    if(b > mid) ret += _Query(a, b, mid + 1, r, (Ind<<1) + 1);    return ret;}inline void _Modify(int a, int l, int r, int Ind, int d){    if(l == r && l == a){        sum[Ind] = d;        return;    }    int mid = (l+r)>>1;    if(a <= mid) _Modify(a, l, mid, Ind<<1, d);    else _Modify(a, mid + 1, r, (Ind<<1) + 1, d);    pushup(Ind);}inline int Query(int a, int b) {return _Query(a, b, 1, size, 1);}inline void Modify(int a, int d){return _Modify(a, 1, size, 1, d);}int main(){    #ifdef LOCAL    freopen("a.txt", "r", stdin);    //freopen("a.out", "w", stdout);    int T = 1;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int n, p, i, u, v, root;    while(cin >> n >> p){        if(n == 0 && p == 0) break;        for(i = 1; i < n; i++){            cin >> u >> v;            sons[u].push_back(v);            sons[v].push_back(u);        }        root = p;        ti = 0;        get_dfs_list(root, -1);        size = ti;        memset(sum, 0, sizeof sum);        for(i = 1; i <= n; i++){            cout << Query(p1[i], p2[i]);            if(i == n) cout << "\n";            else cout << " ";            Modify(p1[i], 1);            sons[i].clear();        }    }    #ifdef LOCAL    cout << endl;    }    #endif // LOCAL    return 0;}

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