PAT_A 1021. Deepest Root (25)

1021. Deepest Root (25)

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

• 分析：题目很简单，但是自己基本功不扎实，做了很长时间。首先，求联通分支，大于１返回ERROR,等于１输出深度最大的节点。这里边最简单的就是查并集，dfs。但是有个缺点是比较耗时，有两个测试点1001ms。做的工程中也看到了别人的代码没有用以上方法，而且耗时比较短。待学习。
• code:学习笔记

#include<iostream>#include<vector>#include<algorithm>using namespace std;int N=0;vector<vector<int> >node;//以后可以写成　vector<int> node[10100];//vector<vector<int> >node;vector<int> v;vector<int> f;vector<int> out;//get father　better way//查并集比较简单//通过查并集确定连通分支还是比较简单的//但是代码比较长，看到别人比较好的代码，//待学习int getF(int a){    if(f[a]==a)return a;    return f[a]=getF(f[a]);}int un(int a,int b){    int fa=0;    int fb=0;    fa=getF(a);    fb=getF(b);    //注意更改父亲时，要更改成父亲的父亲而不是自己的父亲    //不然迭代更新不了,就错乱了    //尽管此时f[a]还没有更新，但可以通过getF()获取更新后的父亲    if(fa>fb)f[fa]=fb;    else if(fa<fb) f[fb]=fa;    if(fa==fb)  return 0;    return 1;}//dfs确定深度，方法很简单，但是容易超时，//在pat上没有超时，但在new code上超时了int dfs(int a){    if(v[a]==1)      return 0;    v[a]=1;    int max=0;    for(int i=0;i<node[a].size();i++)    {        if(v[node[a].at(i)]==0);        {            int tmp=dfs(node[a].at(i));            if(tmp>max)              max=tmp;        }    }    return max+1;}int main(){    cin>>N;    vector<int> tmp_v;    node.assign(N,tmp_v);    f.assign(N,0);    v.assign(N,0);    int tmp=0;    int tmp1=0;    for(int i=0;i<N;i++)      f[i]=i;    for(int i=0;i<N-1;i++)    {        cin>>tmp>>tmp1;        un(tmp1-1,tmp-1);        /*         * 根本不需要的,最初加上这玩意，支持错误，         * 这些工作在un中就已经做了        if(tmp>tmp1)          f[tmp-1]=tmp1-1;        else          f[tmp1-1]=tmp-1;        */        node[tmp-1].push_back(tmp1-1);        node[tmp1-1].push_back(tmp-1);    }    vector<int> ff;    for(int i=0;i<N;i++)        ff.push_back(getF(i));    //去除相同,保证每个元素只有一个    for(int i=0;i<ff.size();i++)    {        for(int j=0;j<i;j++)        {            if(ff.at(i)==ff.at(j))            {              ff.erase(ff.begin()+j);              i--;              break;            }        }    }    if(ff.size()>1)    {        cout<<"Error: "<<ff.size()<<" components"<<endl;    }else    {        int count=0;        for(int i=0;i<N;i++)        {            for(int j=0;j<N;j++)            {                v[j]=0;            }            int tmp=dfs(i);            if(tmp>count)            {                out.clear();                out.push_back(i);                count=tmp;            }else if(tmp==count)            {                out.push_back(i);            }        }        //不用排序，直接输出，因为就是按顺序来查的        for(int i=0;i<out.size();i++)          cout<<out.at(i)+1<<endl;    }    return 0;}

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