[PAT] 1021 Deepest Root (25)（25 分）

1021 Deepest Root (25)（25 分）
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:

5
1 2
1 3
1 4
2 5

Sample Output 1:

3
4
5

Sample Input 2:

5
1 3
1 4
2 5
3 4

Sample Output 2:

Error: 2 components

题意：
给出n个节点（1~n），并给出n-1个边，求图的连通分量。如果连通分量为1，则求图对应的树的最大深度对应的root，如果不唯一则按升序排列输出。

思路：
柳婼 の blog
1.先通过DFS判断图的连通分量个数。
2.任取一个顶点，DFS求得最高高度的结点们，然后从中任取一个结点，再做一个DFS求得最高高度的结点们，做一个并集得到最终结果。

题解：

 #include<cstdio>
 #include<vector>
 #include<set>
 using namespace std;
 vector<vector<int>> mp;
 ];
 ;
 vector<int> temp;
 set<int> s;
 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);
     }
     isVisit[node] = true;
     ; i < mp[node].size(); i++) {
         if (isVisit[mp[node][i]] == false) {
             dfs(mp[node][i], height + );
         }
     }
 }
 int main() {
     int n;
     scanf("%d", &n);
     mp.resize(n + );
     int a, b;
     ; i < n; i++) {
         scanf("%d %d", &a, &b);
         mp[a].push_back(b);
         mp[b].push_back(a);
     }
     ;
     int s1;
     ; i <= n; i++) {
         if (isVisit[i] == false) {
             dfs(i, );
             //假设cnt为1，将temp中的值放入s中。
             //如果cnt>1，这些值虽然不对，但也没用上。
             ) {
                 ) s1 = temp[];
                 ; j < temp.size(); j++) {
                     s.insert(temp[j]);
                 }
             }
             cnt++;
         }
     }
     ) {
         printf("Error: %d components\n", cnt);
     }
     else {
         temp.clear();
         fill(isVisit, isVisit + , false);
         dfs(s1, );
         ; i < temp.size(); i++) {
             s.insert(temp[i]);
         }
         for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
             printf("%d\n", *it);
         }
     }
     ;
 }