1021. Deepest Root (25)

1.采用并查集来检测是否存在2个或2个以上集合

2.使用广度优先搜索进行求多源路径

3.使用Floyd算法会超出内存（可计算10000*10000的int的大小），使用DFS会出现栈溢出（10000深度太深了）

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes' numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print "Error: K components" where K is the number of connected components in the graph.

Sample Input 1:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

//#include<string>//#include <iomanip>#include<vector>#include <algorithm>//#include<stack>#include<set>#include<queue>#include<map>//#include<unordered_set>//#include<unordered_map>//#include <sstream>//#include "func.h"//#include <list>#include<stdio.h>#include<iostream>#include<string>#include<memory.h>#include<limits.h>using namespace std;class VertexNode{public:vector<int> list;VertexNode() :list(0){};};int findRepresent(int a, int* r){if (r[a] == a)return a;else{r[a] = findRepresent(r[a], r);return r[a];}}void dfs(int&node, VertexNode *v, int &maxDeep, int deep, int&root, bool*used, vector<pair<int, int>>&ans){bool visitedAll = true;for (int i = 0; i < v[node].list.size(); i++){int p = v[node].list[i];if (!used[p]){used[p] = true;visitedAll = false;dfs(p, v, maxDeep, deep + 1, root, used, ans);used[p] = false;}}if (visitedAll&&deep >= maxDeep){maxDeep = deep;ans.push_back({ root, deep });}}bool cmp(const pair<int, int>&a, const pair<int, int>&b){if (a.second>b.second)return true;else if (a.second == b.second && a.first < b.first)return true;elsereturn false;}int main(void){int nodeNum;cin >> nodeNum;//VertexNode *v = new VertexNode[nodeNum];//节点vector<VertexNode> v(nodeNum);vector<int> temp(0);for (int i = 0; i < nodeNum - 1; i++){v[i].list = temp;}int *r = new int[nodeNum];//代表for (int i = 0; i < nodeNum; i++)r[i] = i;for (int i = 0; i < nodeNum - 1; i++){int a, b;cin >> a >> b; a--; b--;v[a].list.push_back(b);v[b].list.push_back(a);r[findRepresent(a, r)] = findRepresent(b, r);//合并a和b}int setNum = 0;for (int i = 0; i < nodeNum; i++){//并查集检测集合数量if (i == r[i]) setNum++;}delete[]r;if (setNum >= 2){printf("Error: %d components\n", setNum);return 0;}/*bool *used = new bool[nodeNum];memset(used, false, sizeof(used));*/vector<bool> used (nodeNum, false);vector<pair<int, int>> ans(0);int maxDeep = 0;for (int root = 0; root < nodeNum; root++){//对每一个点使用广度优先搜索,BFS//memset(used, false, sizeof(used));vector<bool> used = vector<bool>(nodeNum, false);queue<int>q;q.push(root);int total = 1;int total2 = 0;int deep = 0;used[root] = true;//cout << "root:" << root << endl;while (!q.empty()){deep++;//cout << "layer" << deep << " :";for (int i = 0; i < total; ++i){int top = q.front(); q.pop();for (int j = 0; j < v[top].list.size(); ++j){int p = v[top].list[j];if (!used[p]){//cout << p << " ";used[p] = true;q.push(p);total2++;}}}total = total2;total2 = 0;//cout << endl;}//cout << endl;if (deep >= maxDeep){maxDeep = deep;ans.push_back({ root, deep });}}sort(ans.begin(), ans.end(), cmp);for (int i = 0; i < nodeNum; i++){used[i] = false;}for (int i = 0; i < ans.size(); i++){if (ans[i].second == maxDeep){used[ans[i].first] = true;//cout << ans[i].first << endl;}else break;}for (int i = 0; i < nodeNum; i++){if (used[i])cout << i + 1 << endl;}return 0;}