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 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components
题目大意:给一个图,首先判断是否为连通图,若非连通图,则分为几个部分?若是连通图,那么以哪些节点为根节点建造的树度最大,输出这几个根节点。
解题思路:读题我们不难发现,判断是否为连通图以及计算以某个节点为根节点的树的度,都可以用DFS解决。首先第一遍DFS判断是否为连通图,思想在之前的题目中也有体现。若为连通图,则我们输出所有达到maxheight的节点。
一开始,我在判断时,以每个结点为根节点都计算了一遍度。。。
还有一个测试点超时了。
for(i=1;i<=n;i++){
fill(visit,visit+10010,false);
dfs(i,1);
}
其实大可不必如此,想想我们在判断是否为连通图的时候,我们就已经得到了以0为根节点,最深的根节点们,想一想,如果我们以这些节点为根节点建树,得到的深度是不是也是最深。所以,只要再以第一遍存储下的节点进行DFS即可,是不是省了很多时间。
#include<iostream>
#include<vector>
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
int n,maxheight=0;
bool visit[10010];
std::vector<vector<int>> graph;
std::vector<int> result;
set<int>node;
void dfs(int node, int height) {
if(height > maxheight) {
result.clear();
result.push_back(node);
maxheight = height;
} else if(height == maxheight){
result.push_back(node);
}
visit[node] = true;
for(int i = 0; i < graph[node].size(); i++) {
if(visit[graph[node][i]] == false)
dfs(graph[node][i], height + 1);
}
}
int numpart(){
int num=0;
for(int i=1;i<=n;i++){
if(visit[i]==false){
num++;
dfs(i,1);
}
}
return num;
}
int main(){
scanf("%d",&n);
graph.resize(n+1);
for(int i=0;i<n-1;i++){
int a,b;
scanf("%d %d",&a,&b);
graph[a].push_back(b);
graph[b].push_back(a);
}
int sum=numpart();
if(sum!=1){
printf("Error: %d components",sum);
}
else{
int r0=result[0];
for(int i=0;i<result.size();i++)
node.insert(result[i]);
result.clear();
maxheight=0;
fill(visit,visit+10010,false);
dfs(r0,1);
for(int i=0;i<result.size();i++)
node.insert(result[i]);
for(auto it=node.begin();it!=node.end();it++)
printf("%d\n",*it);
}
system("pause");
}