1021. Deepest Root (25)

Deepest Root

题目阐述

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

• 原题链接

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题目分析

首先这道题隐含条件非常强！对于N个顶点，给出N-1条边，而且题目强烈的暗示了针对每一个顶点都有一条对应边存在，所以不需要考虑环存在的问题（否则稍微麻烦一些）！然后对于连通分量求算，对于邻接链表逐个进行深搜然后确定即可；对于最大树高度的求算时候，每次深搜过程中保留从出发点计算的当前高度，然后设置全局变量-最大高度，逐个更新即可；参考的柳神的博客，里面解释的非常清楚，然后自己贴上自己实现的代码；

代码

#include <iostream>
#include <vector>
#include <set>
using namespace std;
bool flag[10005];
vector<vector<int>> adj;
vector<int> temp;
int maxHeight = 0;
set<int> s;
void dfs(int u, int height) {
    if (height > maxHeight) {
        temp.clear();
        temp.push_back(u);
        maxHeight = height;
    }
    else if (height == maxHeight) {
        temp.push_back(u);
    }
    flag[u] = true;
    for (int i=0; i<adj[u].size(); i++) {
        if (!flag[adj[u][i]]) dfs(adj[u][i], height+1);
    }
}
int main() {
    int n;
    cin >> n;
    adj.resize(n + 1);
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    int cnt = 0, s1;
    for (int i = 1; i <= n; i++) {
        if (!flag[i]) {
            dfs(i, 1);
            if (i == 1) {
                if (temp.size() != 0) s1 = temp[0];
                for (int j = 0; j < temp.size(); j++)
                    s.insert(temp[j]);
            }
            cnt++;
        }
    }
    if (cnt >=2) {
        printf("Error: %d components", cnt);
        return 0;
    }
    else {
        temp.clear();
        maxHeight = 0;
        fill(flag, flag + 10005, false);
        dfs(s1, 1);
        for (int i = 0; i < temp.size(); i++)
            s.insert(temp[i]);
        for (auto it : s) {
            printf("%d\n", it);
        }
    }
    return 0;
}

总结

欣赏简练优美的代码，就像品尝美味佳酿，让人流连忘返，沉醉不已；