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:

Sample Output 1:

Sample Input 2:

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 ;

 }