1021. Deepest Root (25)

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 calledthe 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

    1. 由于N个节点，N-1条边，则保证如果只有一个连通集，则一定是树，而不可能有环，而连通集个数可以通过一遍DFS得到；

    2. 任取一点DFS，将得到的最远点集合记为s1；

    3. 从s1中任取一点，再来一遍DFS，将得到的最远点集合记为s2；

    4. 合并s1，s2得到s，即为所有根；

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#include <limits>#include <set>#include <functional>using namespace std;vector<vector<bool>> graph;int n;int kcomponents(){    vector<bool> used(n+1, false);    function<void(int)> dfs = [&](int node){        used[node] = true;        for(int i = 1; i <= n; ++i){            if(!used[i] && graph[node][i]) dfs(i);        }    };    int k = 0;    for(int i = 1; i <= n; ++i){        if(!used[i]) dfs(i), ++k;    }    return k;}void DFS(int node, int level, int& maxLevel, vector<int>& nodes, vector<bool>& used){    if(level > maxLevel){        maxLevel = level;        nodes = vector<int>(1, node);    }else if(level == maxLevel){        nodes.push_back(node);    }    for(int i = 1; i <= n; ++i){        if(!used[i] && graph[node][i]){            used[i] = true;            DFS(i, level + 1, maxLevel, nodes, used);            used[i] = false;        }    }}int main(){    scanf("%d", &n);    graph.resize(n+1, vector<bool>(n+1, false));    for(int i = 0; i < n-1; ++i){        int n1, n2;        scanf("%d%d", &n1, &n2);        graph[n1][n2] = graph[n2][n1] = true;    }    int k = kcomponents();    if(k > 1) printf("Error: %d components\n", k);    else{        int maxLevel = numeric_limits<int>::min();        vector<int> nodes;        vector<bool> used(n+1, false);        used[1] = true;        DFS(1, 0, maxLevel, nodes, used);        used[1] = false;        set<int> sets(begin(nodes), end(nodes));        int node = nodes.back();        nodes.clear();        used[node] = true;        maxLevel = numeric_limits<int>::min();        DFS(node, 0, maxLevel, nodes, used);        for(auto& node : nodes){            sets.insert(node);        }        for(auto& node : sets){            printf("%d\n", node);        }    }    return 0;}