标签：des blog http io ar os sp for div
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#
思路如曹博所说，先dfs记录离根的距离，来的方向的前驱。然后按距离排序，之后按此排序求有意义的增加距离。
直观感受是，如果两个叶子在一个分叉上，显然深的节点更先被访问，如果不在一个分叉上，那么先算深的也没什么损失。最后，按照这个权值再对叶子排序一次，就是所要的结果。
#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;
}
标签：des blog http io ar os sp for div
原文地址：http://www.cnblogs.com/lautsie/p/4100869.html
