PAT A1021 Deepest Root （25 分）——图的BFS,DFS

A graph which is connected and acyclic can be considered a tree. The hight 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 (≤10​4​​) 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

作者

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
#include <string>
#include <set>
#include <map>
using namespace std;
const int maxn = ;
const int inf = ;
int n;
int depth[maxn] = {  };
bool vis[maxn] = { false };
struct node {
    int id;
    int depth;
}nodes[maxn];
vector<int> adj[maxn];
void bfs(int v) {
    queue<node> q;
    q.push(nodes[v]);
    vis[v] = true;
    while (!q.empty()) {
        node u = q.front();
        q.pop();
        for (int i = ; i < adj[u.id].size(); i++) {
            if (vis[adj[u.id][i]] == false) {
                nodes[adj[u.id][i]].depth = u.depth + ;
                q.push(nodes[adj[u.id][i]]);
                vis[adj[u.id][i]] = true;
                if (nodes[adj[u.id][i]].depth > depth[v]) {
                    depth[v] = nodes[adj[u.id][i]].depth;
                }
            }
        }
    }
}
bool bfs_c(int v) {
    fill(vis, vis + maxn, false);
    queue<int> q;
    q.push(v);
    vis[v] = true;
    int count = ;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = true;
        for (int i = ; i <adj[u].size(); i++) {
            if (vis[adj[u][i]] == false) {
                q.push(adj[u][i]);
                count++;
                if (count > n)return false;
            }
        }
    }
    return true;
}
int bfsTrave() {
    fill(vis, vis + maxn, false);
    int count = ;
    for (int i = ; i <= n; i++) {
        if (vis[i] == false) {
            bfs(i);
            count++;
        }
    }
    return count;
}
int main() {
    cin >> n;
    for (int i = ; i < n; i++) {
        int c1, c2;
        cin >> c1 >> c2;
        adj[c1].push_back(c2);
        adj[c2].push_back(c1);
    }
    for(int i=;i<=n;i++){
        nodes[i].id = i;
        nodes[i].depth = ;
    }
    int k = bfsTrave();
    if (k > )printf("Error: %d components", k);
    else {
        if (!bfs_c())printf("Error: %d components", k);
        else {
            for (int i = ; i <= n; i++) {
                fill(vis, vis + maxn, false);
                for (int i = ; i <= n; i++) {
                    nodes[i].depth = ;
                }
                bfs(i);
            }
            int max_d = ;
            vector<int> maxi;
            for (int i = ; i <= n; i++) {
                if (depth[i] > max_d) {
                    max_d = depth[i];
                    maxi.clear();
                    maxi.push_back(i);
                }
                else if (depth[i] == max_d) {
                    maxi.push_back(i);
                }
            }
            for (int i = ; i < maxi.size(); i++) {
                printf("%d\n", maxi[i]);
            }
        }
    }
    system("pause");
}

注意点：考察整个图的遍历以及有环无环图的判断。这里判断有没有环我是通过bfs的加入队列个数超过n来判断的。每个节点遍历一遍，找到最大深度再输出。

ps：看了别人的思路，发现自己想多了，n个节点n-1条边，若只有1个联通块就不会有环，所以那个都是白判断的。

ps2：随便找一个节点dfs找到最深的那些节点，再从那些节点里挑一个dfs找到最深的节点，并集就是所有最深的节点，不需要每个节点都做一次搜索。