PAT1021(dfs 连通分量）

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
题目大意：给你一个图，判断这个图能不能构成树，不能构成输出它的连通分量。能构成树，输出以哪些节点为根树的高度最高。
首先通过dfs寻找连通分量的个数，如果不为1，输出无法构成树。
如果为1：两遍dfs找最高的点：
首先以某个点为根，进行dfs()，得到高度最高的点。再从这些点中随机选择一个点再进行dfs，保存高度最高的点，两次遍历的并集即为答案。
代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<set>
using namespace std;
vector<int>v[10010];
bool vis[10010];
vector<int>temp;
set<int>s;
int maxheight;
void dfs(int n,int depth)
{
    if(maxheight<depth)
    {
        maxheight=depth;
        temp.clear();
        temp.push_back(n);
    }
    else if(depth==maxheight)
    {
        temp.push_back(n);
    }
    vis[n]=1;
    for(int i=0;i<v[n].size();i++)
    {
        if(!vis[v[n][i]])
        {
            dfs(v[n][i],depth+1);
        }
    }
}
int main()
{
    int n,a,b;
    scanf("%d",&n);
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n-1;i++)
    {
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int cnt=0,s1;
    for(int i=1;i<=n;i++)
    {
        maxheight=0;
        if(!vis[i])
        {
            dfs(i,1);
            for(int j=0;j<temp.size();j++)
            {
                cout<<temp[j]<<endl;
                s.insert(temp[j]);
                if(j==0)
                    s1=temp[j];
            }
            cnt++;
        }
    }
    if(cnt!=1)
        printf("Error: %d components\n",cnt);
    else
    {
        memset(vis,0,sizeof(vis));
        dfs(s1,1);
        for(int i=0;i<temp.size();i++)
            s.insert(temp[i]);
        set<int>:: iterator it;
        it=s.begin();
        for(it;it!=s.end();it++)
            printf("%d\n",*it);
    }
}