PAT甲级1021

1021. Deepest Root (25)

时间限制

1500 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

#include<cstdio>#include<vector>#include<set>#include<algorithm>using namespace std;const int maxn = 10000;vector<int> G[maxn];bool  vis[maxn] = {false};int maxdepth = 0;int N;set<int> deepestroots;void DFS(int u, int depth){if (maxdepth < depth){maxdepth = depth;deepestroots.clear();deepestroots.insert(u + 1);}else if (maxdepth == depth){deepestroots.insert(u + 1);}vis[u] = true;for (int i = 0; i < G[u].size(); i++){int t = G[u][i];if (!vis[t]){DFS(t, depth + 1);}}}void DFSTrave(vector<int>*G){int count = 0;for (int i = 0; i < N; i++){if (!vis[i]){DFS(i, 1);count++;}}if (count == 1){set<int>::iterator it = deepestroots.begin();int s = *(it)-1;maxdepth = 0;fill(vis, vis + maxn, false);set<int> deepestrootstemp = deepestroots;deepestroots.clear();DFS(s, 1);it = deepestrootstemp.begin();for (it; it != deepestrootstemp.end(); it++){deepestroots.insert(*(it));}it = deepestroots.begin();for (it; it != deepestroots.end(); it++){printf("%d\n", *it);}}else{printf("Error: %d components\n", count);}}//只需至少两次DFS就行，若太多会超时，第一次DFS找出深度最大的那些点，然后从中任选一个进行//第二次DFS，再次找深度最大的哪些点，这两次DFS所找到的点的并集就是，注意去重和排序int main(){scanf("%d", &N);int u, v;if (N == 1){printf("1\n");return 0;}for (int i = 0; i < N - 1; i++){scanf("%d %d", &u, &v);G[u - 1].push_back(v - 1);G[v - 1].push_back(u - 1);}DFSTrave(G);return 0;}