PAT甲级-1021 Deepest Root (25分)

点击链接PAT甲级-AC全解汇总

题目：
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 (≤10^?4?? ) 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：
输入的是一棵树，找树的根使得深度最深
类似于图的最大生成树，但是因为输入的就是树，标记visited比较方便
思路：
第一次dfs，看有几个连通分量，同时找到最深的叶子，这些叶子肯定是可以做最深根的；
第二次dfs，从上一次dfs找到的最深叶子作为根，再找到的最深的叶子是刚才漏的；（因为第一次的起始点可能比较靠近它，所以才会漏，需要第二次dfs找回来）；
两次最深根求并集就行了；

我的代码：

#include<bits/stdc++.h>
using namespace std;
bool node[10010][10010]={
    false};
bool visited[10010]={
    false};
int N,deepest=0;void DFS(int root,int root_depth,set<int>& candidate)
{
    visited[root]=true;if(root_depth>deepest){
    deepest=root_depth;candidate.clear();candidate.insert(root);}else if(root_depth==deepest)candidate.insert(root);for(int next=1;next<=N;next++){
    if(node[root][next]&&!visited[next])DFS(next,root_depth+1,candidate);}
}int main()
{
    cin>>N;for(int i=0;i<N-1;i++){
    int a,b;cin>>a>>b;node[a][b]=true;node[b][a]=true;}//第一次DFS找到最大根候选集1,同时算有几个连通分量int component=0;set<int>candidate1;for(int i=1;i<=N;i++){
    if(visited[i])continue;component++;DFS(i,1,candidate1);}if(component>1){
    printf("Error: %d components",component);return 0;}//第二次DFS找到最大根候选集2fill(visited,visited+10010,false);deepest=0;set<int>candidate2;DFS(*candidate1.begin(),1,candidate2);//两个候选集求并集set<int>result;set_union(candidate1.begin(), candidate1.end(),candidate2.begin(), candidate2.end(),inserter(result, result.begin()));for(set<int>::iterator it=result.begin();it!=result.end();it++){
    cout<<*it<<endl;}return 0;
}