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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

对每个点用BFS求出他的最长路径，效率很低不过还是过了，网上还有2次DFS就可以得出结果，还没有仔细看过，有时间再研究一下

/***并查集求集合数量***//****BFS求最长路径***/#include <iostream>#include <vector>#include <queue>#include <utility>#include <cstring>#define MAX 10005using namespace std;vector<int> graph[MAX];bool visit[MAX];int maxlen[MAX];//记录每个点最大深度int pre[MAX] = {0};int maxh = 0;//记录最大深度int Find(int x){    return pre[x]==x ? x : pre[x]=Find(pre[x]);}void Union(int x, int y){    int u = Find(x);    int v = Find(y);    if(u == v)        return ;    pre[u] = v;}int bfs(int x){    int len = 0;    int dist[MAX] = {0};    queue<int> myque;    myque.push(x);    visit[x] = true;    while(!myque.empty())    {        x = myque.front();        myque.pop();        visit[x] = true;        for(vector<int>::size_type i=0; i<graph[x].size(); i++)        {            if(!visit[graph[x][i]])            {                dist[graph[x][i]] = dist[x]+1;                if(len < dist[graph[x][i]])                    len = dist[graph[x][i]];                myque.push(graph[x][i]);            }        }    }    if(len > maxh)        maxh = len;    return len;}int main(){    int n, x, y;    int maxl = -1;    int counter = 0;    cin >> n;    for(int i=0; i<=n; i++)        pre[i] = i;    for(int i=0; i<n-1; i++)    {        cin >> x >> y;        graph[x].push_back(y);        graph[y].push_back(x);        Union(x, y);    }    for(int i=1; i<=n; i++)    {        if(pre[i] == i)            counter++;    }    if(counter > 1)        cout << "Error: " << counter <<" components" << endl;    else    {        for(int i=1; i<=n; i++)        {            memset(visit, false, sizeof(bool)*(n+1));            maxlen[i] = bfs(i);        }        for(int i=1; i<=n; i++)        {            if(maxlen[i] == maxh)                cout << i << endl;        }    }    return 0;}