1021 Deepest Root (25)(25 分)
题目描述:
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.
输入格式:
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.输出格式:
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.
题目大意:
题目就是要对一个树形结构,去寻找能够以某个结点以其为根,深度最大,如果有多个,则升序输出。解题思路参考自这里
解题方法:
采用dfs对图进行遍历,因为如果是连通图,dfs就能把所有结点都遍历完,而如果不是连通图,则需要至少2次以上,因此根据这个特性可以用来判断是否是树。同时,通过dfs,可以将最深的叶子结点找出来。这里需要注意一点,类似于(1-2-3)这样的树形结构,1和3都可以被称为root结点,因此,我们可以通过对第一遍dfs得到遍历结果中,随便取一个,反向dfs,最终把结果全部放入set中,把两次dfs的并集放入。
程序:
#include <stdio.h>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
int maxHight = 0;
set <int> s; /* 保存深度最大的根结点集合(默认升序) */
vector <vector<int> > v; /* 保存点与点之间的联系 */
vector <int> temp; /* 保存深度最大点的向量 */
bool visit[10001];
void dfs(int node, int height)
{ /* 深度优先搜索遍历 */
if (height > maxHight) /* 如果当前结点的深度比最大深度还要大 */
{
maxHight = height;
temp.clear();
temp.push_back(node);
}
else if (height == maxHight) /* 如果相同,将相同深度的结点放入temp */
temp.push_back(node);
visit[node] = true; /* 置该点为已访问 */
for (int i = 0; i < v[node].size(); i++) /* 遍历v[node]的所有孩子 */
if (visit[v[node][i]] == false)
dfs(v[node][i], height+1);
}
int main(int argc, char const *argv[])
{
int N, father, child, cnt = 0;
scanf("%d", &N);
fill(visit, visit+N+1, false); /* 初始化visit */
v.resize(N+1); /* 调整vector的大小 */
for (int i = 0; i < N-1; i++) /* 输入父节点和孩子节点之间的联系 */
{
scanf("%d %d", &father, &child);
v[father].push_back(child);
v[child].push_back(father);
}
for (int i = 1; i <= N; i++) /* 编号从1开始 */
if (visit[i] == false) /* 如果还未访问过 */
{
dfs(i, 1);
if (i == 1)
for (int j = 0; j < temp.size(); j++) /* 把当前的叶子结点存入 */
s.insert(temp[j]);
cnt++; /* 每执行一次cnt++,用于判断是否是树 */
}
if (cnt >= 2) /* 如果不是树 */
printf("Error: %d components\n", cnt);
else /* 从叶子结点出发,去寻找最深的根结点 */
{
fill(visit, visit+N+1, false); /* 重置visit */
temp.clear();
maxHight = 0;
dfs(temp[0], 1); /* 因为树是连通的,所以从任意一个叶节点出发都能遍历到所有最深的父节点 */
for (int j = 0; j < temp.size(); j++)
s.insert(temp[j]);
set <int>::iterator it;
for (it = s.begin(); it != s.end(); it++)
printf("%d\n", *it);
}
return 0;
}
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