PAT 甲级 1021 Deepest Root (25 分)（bfs求树高,又可能存在part数part>2的情况）

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。

5
1 2
1 3
3 1
2 3
Error: 3 components

 

AC代码：

#include<bits/stdc++.h> 
using namespace std;
int n;
vector<int>v[10005];
struct node{
    int k;//节点的值 
    int h;//在第几层 
};
queue<node>q;
queue<int>ans;
int in[100005];//在不在队列里 
int main(){
    cin>>n;
    for(int i=1;i<=n-1;i++){
        v[i].clear();
    }
    while(!q.empty()) q.pop();
    while(!ans.empty()) ans.pop();
    for(int i=1;i<=n-1;i++){
        int x,y;
        cin>>x>>y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    int maxH=0;
    memset(in,0,sizeof(in));
    int part=1; 
    for(int i=1;i<=n;i++)
    {
        //以i为树根
        int height=0;
        memset(in,0,sizeof(in));
        node x;
        x.k=i;
        x.h=1;
        q.push(x);
        in[i]=1;//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=0;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+1;//新的一层高度+1再放进队列 
                q.push(y);
                in[k1]=1;            
            }            
        }
        //先检查是不是一个块的
        for(int j=1;j<=n;j++)
        {
            if(in[j]!=1)
            {
                part=0;
                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=1;
        for(int j=1;j<=n;j++)
        {//bfs数数有多少块 
            if(in[j]!=1)
            {
                part++;//块数+1 
                node x;
                x.k=j;
                x.h=1;
                q.push(x);
                in[j]=1;
                while(!q.empty())
                {
                    node x=q.front();
                    q.pop();
                        for(int p=0;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+1;
                            q.push(y);
                            in[k1]=1;            
                        }
                    }
            }
        }
        cout<<"Error: "<<part<<" components";
    }else{
        while(!ans.empty()){
            cout<<ans.front()<<endl;
            ans.pop();
        }
    }
    return 0;    
}