PAT 1021 Deepest Root[并查集、dfs][难]

1021 Deepest Root (25)（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 (<=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:对于图如果其是联通无环的，那么就是树，找出最深的树的所有起点。

 //不知道怎么找到所有的，找到两个可还行。

//使用并查集判断是否是树，然后呢？

代码来自:https://www.liuchuo.net/archives/2348

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int n, maxheight = 0;
vector<vector<int>> v;
bool visit[10010];
set<int> s;
vector<int> temp;
void dfs(int node, int height) {
    if(height > maxheight) {
        temp.clear();//只存放最深的。
        temp.push_back(node);
        maxheight = height;
    } else if(height == maxheight){
        temp.push_back(node);
    }
    visit[node] = true;//一般图访问点完了之后，都要标记的，防止其下一个点再返回去访问它。
    for(int i = 0; i < v[node].size(); i++) {
        if(visit[v[node][i]] == false)
            dfs(v[node][i], height + 1);
    }
}
int main() {
    scanf("%d", &n);
    v.resize(n + 1);
    int a, b, cnt = 0, s1 = 0;
    for(int i = 0; i < n - 1; i++) {
        scanf("%d%d", &a, &b);
        v[a].push_back(b);//使用二维向量来存储。
        v[b].push_back(a);
    }
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false) {
            dfs(i, 1);
            if(i == 1) {//其实这里随便一个点均可。
                if (temp.size() != 0) s1 = temp[0];
                for(int j = 0; j < temp.size(); j++)
                    s.insert(temp[j]);//再使用集合存储，防止重复。
            }
            cnt++;
        }
    }
    if(cnt >= 2) {
        printf("Error: %d components", cnt);
    } else {
        temp.clear();
        maxheight = 0;
        fill(visit, visit + 10010, false);
        dfs(s1, 1);
        for(int i = 0; i < temp.size(); i++)
            s.insert(temp[i]);
        for(auto it = s.begin(); it != s.end(); it++)
            printf("%d\n", *it);
    }
    return 0;
}

 

//厉害，使用dfs，传递层数参数，并且有一个maxHeight来保存最大的，厉害。