A1021. 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

给出n个结点（1~n）之间的n条边，问是否能构成一棵树，如果不能构成则输出它有的连通分量个数，如果能构成一棵树，输出能构成最深的树的高度时，树的根结点。如果有多个，按照从小到大输出。

#include <cstdio>#include <vector>#include <set>#include <algorithm>using namespace std;int n, maxheight = 0;vector<vector<int>> v;//二维数组，路ab，用v[a][i]表示路,值为路顶点的对端b,i是顶点引出来的第几条路bool visit[10010];set<int> s;//输出结果vector<int> temp;//记录暂时最高的点void dfs(int node, int height) {    if(height > maxheight) {        temp.clear();        temp.push_back(node);        maxheight = height;    } else if(height == maxheight){        temp.push_back(node);    }    visit[node] = true;    for(int i = 0; i < v[node].size(); i++) {        if(visit[v[node][i]] == false)//如果该顶点对端未被访问，则访问之            dfs(v[node][i], height + 1);    }}int main() {    scanf("%d", &n);    v.resize(n + 1);    int a, b, cnt = 0;//cnt连通分量    for(int i = 0; i < n - 1; i++) {        scanf("%d%d", &a, &b);        v[a].push_back(b);        v[b].push_back(a);    }    int s1 = 0; //同样最大高度的第一个根结点    for(int i = 1; i <= n; i++) {        if(visit[i] == false) {            dfs(i, 1);            if(i == 1) {                for(int j = 0; j < temp.size(); j++) {                    s.insert(temp[j]);                    if(j == 0) s1 = temp[j];                }            }            cnt++;        }    }    if(cnt >= 2) {        printf("Error: %d components", cnt);    } else {        temp.clear();        maxheight = 0;        fill(visit, visit + 10010, false);        dfs(s1, 1);        for(int i = 0; i < temp.size(); i++)            s.insert(temp[i]);        for(set<int>::iterator it = s.begin(); it != s.end(); it++)            printf("%d\n", *it);            }    return 0;}