*[codility]Country network

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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#

思路如曹博所說，先dfs記錄離根的距離，來的方向的前驅。然後按距離排序，之後按此排序求有意義的增加距離。

直觀感受是，如果兩個葉子在一個分叉上，顯然深的節點更先被訪問，如果不在一個分叉上，那麼先算深的也沒什麼損失。最後，按照這個權值再對葉子排序一次，就是所要的結果。

#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