PTA (Advanced Level) 1021 Deepest Root

Deepest Root

　　A graph which is connected and acyclic can be considered a tree. The hight 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

题目解析
　　本题给出一个无向图的信息，包括一个数字n，代表无向图有n个结点，之后跟随n-1行分别为n-1条边连接的两个结点。要求判断该无向图是否是一颗树，若是一棵树，由小到大输出以其为根结点时深度最大的点，若不是树则输出连通块数量。

　　在解题之前首先看一下题目的限制

　　时间限制: 2000 ms

　　内存限制: 64 MB

　　

　　本题结点数量最多1e4个若用二维数组存储邻接矩阵，虽然数组没有超限但是当节点数取到最大1e4个结点时，需要的内存空间超过382MB，所以在这里使用邻接表来存储无向图。对于判断无向图是否为树，由于无向图有n个结点n-1条边，那么只要这个图是连通图，它便肯定是一颗树。我们只需要维护一个并查集便可以达到判断数与输出连通块数量，由于本题时间限制非常松，时间宽松使人暴力，所以在寻找深度最大的根结点时，这里以所有点为根结点暴力深搜获取深度，用set保存所有深度最大的点，之后将其输出即可。

 #include<bits/stdc++.h>
 using namespace std;
 const int MAX =1e4 + ;
 vector<int> G[MAX];
 int f[MAX];
 int n;
 void init(){    //初始化并查集
     for(int i = ; i <= n; i++)
         f[i] = i;
 }
 int getF(int x) //并查集找爹函数
 {
     if(f[x] == x)
         return x;
     else
         return f[x] = getF(f[x]);
 }
 int maxH = ;
 void DFS_getH(int u, int height, int preNode){  //深搜获取深度
     maxH = max(maxH, height);
     for(int i = ; i < G[u].size(); i++)
         if(G[u][i] != preNode)
             DFS_getH(G[u][i], height + , u);
 }
 int main(){
     scanf("%d", &n);    //输入结点数量
     init(); //初始化并查集
     memset(G, , sizeof(G));
     for(int i = ; i < n - ; i++){ //输入n-1条边
         int u, v;
         scanf("%d%d", &u, &v);
         G[u].push_back(v);
         G[v].push_back(u);
         int fu = getF(u);
         int fv = getF(v);
         if(fu != fv)    //合并一条的两个端点
             f[fu] = fv;
     }
     int cnt = ;    //记录连通块个数
     for(int i = ; i <= n; i++){    //获取连通块个数
         if(f[i] == i)
             cnt++;
     };
     if(cnt != )    //若连通块个数不等于1证明给出的图不是一颗树
         printf("Error: %d components\n", cnt);   //输出连通块个数
     else{
         set<int> ans;
         int ansH = ;
         for(int i = ; i <= n; i++){
             maxH = ;
             DFS_getH(i, , );  //获取每一个点为根时对应的树高（用maxH记录）
             if(maxH > ansH){    //ansH记录当前最高深度
                 ansH = maxH;    //若找到深度更大的点便清空集合重新记录
                 ans.clear();
                 ans.insert(i);
             }else if(maxH == ansH){ //找到深度与当前最大深度相同的点便加入集合
                 ans.insert(i);
             }
         }
         for(auto i : ans){
             printf("%d\n", i);
         }
     }
     return ;
 }