*[codility]country Network

https://codility.com/programmers/challenges/fluorum2014

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1273

http://blog.csdn.net/caopengcs/article/details/36872627

http://www.quora.com/ how-do-i-determine-the-order-of-visiting-all-leaves-of-a-rooted-tree-so-that-in-each-step-i-visit-a-leaf-whose-path-from- root-contains-the-most-unvisited-nodes#

Thinking as Cao Bo said, first DFS records the distance from the root, the direction of the precursor. Then sort by distance, then sort by this to make a meaningful increase in distance.

The intuitive feeling is that if the two leaves on a fork, it is clear that the deep node is first accessed, if not a fork, then the first depth is no loss. Finally, sort the leaves again according to this weight, which is the desired result.

#include <iostream>using namespace Std;void dfs (vector<vector<int>> &tree, vector<int>    &parent, vector<int> &depth, vector<bool> &visited, int root, int dep) {Visited[root] = true;    Depth[root] = DEP;        for (int i = 0; i < tree[root].size (); i++) {if (Visited[tree[root][i]]) continue;        Parent[tree[root][i]] = root;    DFS (tree, parent, depth, visited, Tree[root][i], dep + 1);    }}vector<int> solution (int K, vector<int> &t) {int n = t.size ();    Vector<vector<int>> tree (n);            for (int i = 0; i < n; i++) {if (t[i]! = i) {tree[i].push_back (t[i]);        Tree[t[i]].push_back (i);    }} vector<int> parent (n);    Vector<int> depth (n);    Vector<bool> visited (n);    DFS (tree, parent, depth, visited, K, 0);    Vector<vector<int>> CNT (n); for (int i = 0; i < n; i++) {Cnt[depth[i]].push_back (i);    } vector<int> ordered; for (int i = n-1; I >= 0, i--) {for (int j = 0; J < Cnt[i].size (); j + +) {Ordered.push_back (CNT I            [j]);    }} vector<int> Res;    Vector<int> length (n);    Visited.clear ();    Visited.resize (n);    Visited[k] = true;        for (int i = 0; i < ordered.size (); i++) {int len = 0;        int x = Ordered[i];            while (!visited[x]) {visited[x] = true;            x = Parent[x];        len++;        } Length[ordered[i]] = len;    cout << "length[" << Ordered[i] << "]:" << len << Endl;    } cnt.clear ();    Cnt.resize (n);        for (int i = 0; i < length.size (); i++) {cnt[length[i]].push_back (i);    } res.push_back (K); for (int i = Cnt.size ()-1, i > 0; i--) {for (int j = 0, J < Cnt[i].size (); j + +) {Res.push_back        (Cnt[i][j]); }} return res;}

　　

*[codility]country Network