PAT

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

---

给定条件：
1.n个节点
2.n-1条连线

题目要求：
1.判断给定的图是否连通的
2.如果连通，求出选取那些节点作为根节点，所建成的树的深度最大
3.按照编号从小到大的顺序，输出符合条件的根节点

解决方法：
1.dfs求图的连通分量，判断是否为1
2.分别选取所有的节点作为根节点，进行深度优先遍历，记录所到达的最大深度，并且在搜索过程中记录每个节点所能处于的最大深度
3.遍历所有节点，输出所在深度可以为最大深度的节点

---

#include<cstdio>#include<vector>using namespace std;int N, a, b, maxd = -1;int vis[10001];int depth[10001];vector<int> e[10001];void dfs(int v, int d){vis[v] = 1;depth[v] = max(depth[v], d); // update node's depthmaxd = max(maxd, d); // update the max depthfor(int i = 0; i < e[v].size(); i++){int next = e[v][i];if(!vis[next]){dfs(next,d+1);}}}int main(){while(scanf("%d", &N) != EOF){// inputfor(int i = 1; i < N; i++){scanf("%d%d", &a, &b);e[a].push_back(b);e[b].push_back(a);}fill(vis, vis+10001, 0);int cnt = 0;for(int i = 1; i <= N; i++){if(!vis[i]){dfs(i,1);cnt++; // the number of components}}if(cnt != 1){printf("Error: %d components\n",cnt);}else{// dfs every nodefor(int i = 1; i <= N; i++){fill(vis, vis+10001, 0);vis[i] = 1;dfs(i,1);}// find the node which depth is maxdfor(int i = 1; i <=N; i++){if(depth[i] == maxd){printf("%d\n", i);}}}}return 0;}