[PAT] 1021 Deepest Root (25)（25 分）

1021 Deepest Root (25)（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 (<=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:

Sample Output 1:

3
4
5

Sample Input 2:

Sample Output 2:

Error: 2 components

题意：
给出n个节点（1~n），并给出n-1个边，求图的连通分量。如果连通分量为1，则求图对应的树的最大深度对应的root，如果不唯一则按升序排列输出。

思路：
柳婼の blog
1.先通过DFS判断图的连通分量个数。
2.任取一个顶点，DFS求得最高高度的结点们，然后从中任取一个结点，再做一个DFS求得最高高度的结点们，做一个并集得到最终结果。

题解：

 1 #include<cstdio>
 2 #include<vector>
 3 #include<set>
 4 using namespace std;
 5 vector<vector<int>> mp;
 6 bool isVisit[10010];
 7 int maxHeight = 1;
 8 vector<int> temp;
 9 set<int> s;
10 void dfs(int node, int height) {
11     if (height > maxHeight) {
12         temp.clear();
13         temp.push_back(node);
14         maxHeight = height;
15     }
16     else if (height == maxHeight) { 
17         temp.push_back(node);
18     }
19     isVisit[node] = true;
20     for (int i = 0; i < mp[node].size(); i++) {
21         if (isVisit[mp[node][i]] == false) {
22             dfs(mp[node][i], height + 1);
23         }
24     }
25 }
26 int main() {
27     int n;
28     scanf("%d", &n);
29     mp.resize(n + 1);
30     int a, b;
31     for (int i = 1; i < n; i++) {
32         scanf("%d %d", &a, &b);
33         mp[a].push_back(b);
34         mp[b].push_back(a);
35     }
36     int cnt = 0;
37     int s1;
38     for (int i = 1; i <= n; i++) {
39         if (isVisit[i] == false) {
40             dfs(i, 1);
41             //假设cnt为1，将temp中的值放入s中。
42             //如果cnt>1，这些值虽然不对，但也没用上。
43             if (i == 1) {
44                 if (temp.size() != 0) s1 = temp[0];
45                 for (int j = 0; j < temp.size(); j++) {
46                     s.insert(temp[j]);
47                 }
48             }
49             cnt++;
50         }
51     }
52     if (cnt >= 2) {
53         printf("Error: %d components\n", cnt);
54     }
55     else {
56         temp.clear();
57         fill(isVisit, isVisit + 10010, false);
58         dfs(s1, 1);
59         for (int i = 0; i < temp.size(); i++) {
60             s.insert(temp[i]);
61         }
62         for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
63             printf("%d\n", *it);
64         }
65     }
66     return 0;
67 }

[PAT] 1021 Deepest Root (25)（25 分）

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