1021 Deepest Root (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 (≤) 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

题意:

给出n个结点和n-1条边,问它们能否形成一棵n个结点的树,如果能,从中选出结点作为树根,使整棵树的高度最大。按升序输出所有满足要求的可以作为树根的结点。 如果不是一棵树,则输出cout<<"Error: "<<part<<" components";

思路:

bfs求高度,如果有多个部分,再bfs数有几个部分,part可能大于2。一开始没考虑,测试点2过不了。


Error:  components

AC代码:

#include<bits/stdc++.h>
using namespace std;
int n;
vector<int>v[];
struct node{
int k;//节点的值
int h;//在第几层
};
queue<node>q;
queue<int>ans;
int in[];//在不在队列里
int main(){
cin>>n;
for(int i=;i<=n-;i++){
v[i].clear();
}
while(!q.empty()) q.pop();
while(!ans.empty()) ans.pop();
for(int i=;i<=n-;i++){
int x,y;
cin>>x>>y;
v[x].push_back(y);
v[y].push_back(x);
}
int maxH=;
memset(in,,sizeof(in));
int part=;
for(int i=;i<=n;i++)
{
//以i为树根
int height=;
memset(in,,sizeof(in));
node x;
x.k=i;
x.h=;
q.push(x);
in[i]=;//标记已访问
while(!q.empty())
{
node x=q.front();
q.pop();
height=max(height,x.h);//更新高度
for(int j=;j<v[x.k].size();j++)//遍历与 x.k相连的节点
{
int k1=v[x.k].at(j);
if(in[k1])//被访问过了
{
continue;
}
node y;
y.k=k1;
y.h=x.h+;//新的一层高度+1再放进队列
q.push(y);
in[k1]=;
}
}
//先检查是不是一个块的
for(int j=;j<=n;j++)
{
if(in[j]!=)
{
part=;
break;
}
}
if(!part)
{
break;
}
//cout<<i<<" "<<height<<endl;
//更新高度
if(height>maxH)
{
maxH=height;
while(!ans.empty()) ans.pop();//更新了就清空
ans.push(i);
}else if(height==maxH)
{
ans.push(i);
}
}
if(!part){
part=;
for(int j=;j<=n;j++)
{//bfs数数有多少块
if(in[j]!=)
{
part++;//块数+1
node x;
x.k=j;
x.h=;
q.push(x);
in[j]=;
while(!q.empty())
{
node x=q.front();
q.pop();
for(int p=;p<v[x.k].size();p++)
{
int k1=v[x.k].at(p);
if(in[k1])
{
continue;
}
node y;
y.k=k1;
y.h=x.h+;
q.push(y);
in[k1]=;
}
}
}
}
cout<<"Error: "<<part<<" components";
}else{
while(!ans.empty()){
cout<<ans.front()<<endl;
ans.pop();
}
}
return ;
}
 

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