1021 Deepest Root (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 (≤) 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

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 10005  //卡4088我也是醉了

#define INF 9999999

int G[MAXN][MAXN];
int vis[MAXN] = {0};
int n;
int maxdepth = 0;
vector<vector<int>> mp;

set<int> st;
vector<int> vec;

void dfs(int x,int depth){
    if(depth > maxdepth){vec.clear();vec.push_back(x);maxdepth = depth;}    //这个写法在图中很常见，用vec来保存最长的路径
    else if(depth == maxdepth){vec.push_back(x);}
    vis[x] = 1;
    for(int i=0;i < mp[x].size();i++){
        if(!vis[mp[x][i]]){
            dfs(mp[x][i],depth+1);
        }
    }
}



int main(){

    cin >> n;

    mp.resize(n+1);   //要先开辟空间，不然下面赋值不了
    for(int i=1;i <= n-1;i++){
        int x1,y1;
        cin >> x1 >> y1;
        mp[x1].push_back(y1);
        mp[y1].push_back(x1);
    }


    int cnt=0;
    for(int i=1;i <= n;i++){
        if(vis[i] == 0){
            dfs(i,1);
            cnt++;
        }
    }
    if(cnt!=1){
        printf("Error: %d components",cnt);
        return 0;
    }

    int x = (vec.size())?vec[0]:0;

    fill(vis,vis+n+1,0); //1-n这里一开始写错了
    for(int i=0;i < vec.size();i++){
        st.insert(vec[i]);
    }
    vec.clear();maxdepth = 0;

    dfs(x,1);

    for(int i=0;i < vec.size();i++){
        st.insert(vec[i]);
    }

    for(auto it=st.begin();it!=st.end();it++){
        cout << *it << endl;
    }

    return 0;
}

卡内存，要用vector不用二维数组

先dfs一次找到离1（随便一个节点）最远的节点保存再vec中

然后再从对任意一个vec中的节点dfs找到相对最远的节点，然后对两次的dfs取交集

尼玛好难啊透