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 1:5 1 2 1 3 1 4 2 5Sample Output 1:
3 4 5Sample Input 2:
5 1 3 1 4 2 5 3 4Sample Output 2:
Error: 2 components
#include <iostream>
#include <map>
#include <vector>
using namespace std;
int n;
int a[10001][10001];
int mark[10001];
int components = 0;
int maxLevel = 0;
vector<int> elements;
void dfs(int root, int i, int level) {
if (level > maxLevel) {
maxLevel = level;
elements.clear();
elements.push_back(root);
}
else if (level == maxLevel) {
if (root != elements[elements.size() - 1])
elements.push_back(root);
}
for (int j = 1; j <= n; j++) {
if (a[j][i] > 0 && j != i&&mark[j] == 0) {
mark[j] = 1;
dfs(root, j, level + 1);
}
}
}
void CalcComp(int p) {
for (int i = 1; i <= n; i++) {
if (a[p][i] > 0 && p != i&&mark[i] == 0) {
mark[i] = 1;
CalcComp(i);
}
}
}
int main(void) {
cin >> n;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
a[u][v] = 1;
a[v][u] = 1;
}
while (true)
{
int p, cnt = 0;
for (int i = 1; i <= n; i++) {
if (mark[i] == 0) {
components++;
mark[i]++;
p = i;
break;
}
else {
cnt++;
}
}
if (cnt == n) {
cnt = 0;
break;
}
CalcComp(p);
}
if (components > 1) {
cout << "Error: " << components << " components";
return 0;
}
if (n <= 15) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
mark[j] = 0;
mark[i] = 1;
dfs(i, i, 1);
}
}
else {
for (int i = 1; i <= n; i++) {
int cnt = 0;
for (int j = 1; j <= n; j++) {
cnt += a[i][j];
}
if (cnt == 1)
elements.push_back(i);
cnt = 0;
}
}
for (int i = 0; i < elements.size(); i++) {
cout << elements[i] << endl;
}
return 0;
}
原文:http://www.cnblogs.com/zzandliz/p/5023075.html