1021. Deepest Root (25)

题目连接：https://www.patest.cn/contests/pat-a-practise/1021

原题如下：

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

这道题和1013很像，1013是求出连通集的个数即可，这道题不仅得利用连通集个数，还得求最大深度。
我一开始想的有点乱，尤其是求最大深度这块。其实这块代码的和求树的高度类似……（还是递归学得不扎实啊）。因为参考了别人的代码，图的存储用vector，
其实用数组也行。
另外，题中的条件“有N个点，N-1条边”也很有用 －> 如果树中有环，一定会ERROR
总之，太乱了……后面有时间觉得还是再写一遍吧。

#include<iostream>
#include<vector>
#include<string.h>
#include<stdio.h>
#define MAXN 10005
using namespace std;
vector<int>Edge[MAXN];
int deep[MAXN],visited[MAXN]={0};
int N;

int DFS(int v)
{

    int i,tmp=0,Maxtmp=0;
    visited[v]=1;
    for (i=0;i<Edge[v].size();i++)
    {
        if (!visited[Edge[v][i]])tmp=DFS(Edge[v][i]);

        if (tmp>Maxtmp)Maxtmp=tmp;
    }
    return Maxtmp+1;
}

int main()
{
    scanf("%d",&N);
    int v,w,i,j;
    for (i=0;i<N-1;i++)
        {scanf("%d %d",&v,&w);
        Edge[v].push_back(w);
        Edge[w].push_back(v);
        }

    memset(deep,0,sizeof(deep));

    int tag=0,Maxdeep=0;

    for (i=1;i<=N;i++)
    {
        if(tag==0){memset(visited,0,sizeof(visited));
        deep[i]=DFS(i);
        Maxdeep=Maxdeep>deep[i]?Maxdeep:deep[i];}

        for (j=1;j<=N;j++)
        {
            if (!visited[j])
            {
                tag++;
                i=j-1;
                break;
            }
        }
    }

   if (tag>0)
    {
        printf("Error: %d components",tag+1);
    }
    else
    {
        for (i=1;i<=N;i++)
        {
            if (deep[i]==Maxdeep)printf("%d\n",i);
        }
    }
    return 0;
}