1021 Deepest Root （PAT甲级）-Toy模板网

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1021. Deepest Root (25)-PAT甲级真题（图的遍历，dfs，连通分量的个数）_柳婼的博客-CSDN博客

柳婼的解法在这里，两次dfs，还是挺好玩的。

我的解法比较暴力，就是先用并查集算连通分量（这个其实还是dfs来算会更方便），如果只有一个连通分量，那deepest root一定在仅有一条arc的点上（除了只有一个结点的树，可以一开始就特殊考虑），然后从这些点开始dfs，算出深度最高的那些root。

#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
const int maxN = 10001;

int N, u, v, maxx, layer, maxL;
int fa[maxN];
std::set<int> st, root;
std::vector<std::vector<int>> adj;
std::vector<bool> visited;

int findFather(int m){
    int a = m;
    while(fa[a] != a){
        a = fa[a];
    }
    int t;
    while(fa[m] != m){
        t = fa[m];
        fa[m] = a;
        m = t;
    }
    return a;
}

void myUnion(int m, int n){
    fa[findFather(m)] = findFather(n);
}

void dfs(int k){
    visited[k] = true;
    ++layer;
    if(layer > maxL){
        maxL = layer;
    }
    for(int i = 0; i < adj[k].size(); ++i){
        if(!visited[adj[k][i]]){
            dfs(adj[k][i]);
        }
    }
    --layer;
}

int main(){
    scanf("%d", &N);
    if(N == 1){
        printf("1");
        return 0;
    }
    adj.resize(N + 1);
    visited.resize(N + 1);
    for(int i = 1; i <= N; ++i){
        fa[i] = i;
    }
    for(int i = 1; i < N; ++i){
        scanf("%d %d", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
        myUnion(u, v);
    }
    for(int i = 1; i <= N; ++i){
        st.insert(findFather(i));
    }
    if(st.size() != 1){
        printf("Error: %d components", st.size());
        return 0;
    }
    maxx = 0;
    for(int i = 1; i <= N; ++i){
        if(adj[i].size() == 1){
            maxL = 0;
            layer = 0;
            std::fill(visited.begin() + 1, visited.end(), false);
            dfs(i);
            if(maxL > maxx){
                maxx = maxL;
                root.clear();
                root.insert(i);
            } else if(maxL == maxx){
                root.insert(i);
            }
        }
    }
    for(auto c : root){
        printf("%d\n", c);
    }
    return 0;
}

题目如下：

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 (≤104) 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 componentswhere K is the number of connected components in the graph.文章来源地址https://www.toymoban.com/news/detail-614186.html