PTA数据结构-06-图1 列出连通集

一、题目

给定一个有N个顶点和E条边的无向图，请用DFS和BFS分别列出其所有的连通集。假设顶点从0到N−1编号。进行搜索时，假设我们总是从编号最小的顶点出发，按编号递增的顺序访问邻接点。

输入格式:

输入第1行给出2个整数N(0

输出格式:

按照"{ v​1​​ v​2​​ ... v​k​​ }"的格式，每行输出一个连通集。先输出DFS的结果，再输出BFS的结果。

输入样例:

8 6
0 7
0 1
2 0
4 1
2 4
3 5

输出样例:

{ 0 1 4 2 7 }
{ 3 5 }
{ 6 }
{ 0 1 2 7 4 }
{ 3 5 }
{ 6 }

 

二、解答

import java.util.*;

public class Main{
    static ArrayList stack = new ArrayList<>();
    static int index = -1;
    static ArrayList queue = new ArrayList<>();
    static int front = -1, rear = -1;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int[][] map = creatMap(sc);
        printAllRes(map);
    }
    public static void printAllRes(int[][] map){
        int nodeNum = map.length;
        int[] visited = new int[nodeNum];
        for (int i = 0; i < nodeNum; i++) {
            if (visited[i] == 0) System.out.println(DFS(i, visited, map));
        }
        visited = new int[nodeNum];
        for (int i = 0; i < nodeNum; i++) {
            if (visited[i] == 0) System.out.println(BFS(i, visited, map));
        }

    }
    public static String DFS(int node, int[] visited, int[][] map){
        String res = "";
        int nodeNum = map.length;
        while(node != -1) {
            //某个点可能与多个点相邻，被多次压入堆栈
            if(visited[node] == 1) { node = pop(); continue;}
            //从小到大输出，因此避免倒序
            for (int i = nodeNum - 1; i >= 0; i--) {
                if (map[node][i] != 0 && visited[i] == 0) push(i);
            }
            visited[node] = 1;
            res += node + " ";
            node = pop();
        }
        return "{ " + res.trim() + " }";
    }
    public static String BFS(int node, int[] visited, int[][] map){
        String res = "";
        int nodeNum = map.length;
        while(node != -1) {
            //某个点可能与多个点相邻，被多次压入堆栈
            if(visited[node] == 1) { node = outQue(); continue;}
            for (int i = 0; i < nodeNum; i++) {
                if (map[node][i] != 0 && visited[i] == 0) inQue(i);
            }
            visited[node] = 1;
            res += node + " ";
            node = outQue();
        }
        return "{ " + res.trim() + " }";
    }

    public static int[][] creatMap(Scanner sc){
        int nodeNum = sc.nextInt();
        int lineNum = sc.nextInt();
        int [][] map = new int[nodeNum][nodeNum];
        for (int i = 0; i < lineNum; i++) {
            int node1 = sc.nextInt();
            int node2 = sc.nextInt();
            map[node1][node2] = 1;
            map[node2][node1] = 1;
        }
        return map;
    }
    public static void push(int num){
        stack.add(num);
        index++;
    }
    public static int pop(){
        if (index < 0) return -1;
        return stack.remove(index--);
    }
    public static void inQue(int num){
        queue.add(num);
        front++;
    }
    public static int outQue(){
        if (rear >= front) return -1;
        return queue.get(++rear);
    }
}

思路：

1. 考虑图的保存和遍历：二维数组
2. DFS：深度遍历过程，将邻近点压入堆栈
3. BFS：宽度遍历过程，将邻近点压入队列
4. 某个点可能与多个点相邻，被多次压入堆栈，只处理一次

思考：

1. 很多时候递归可以用堆栈来解决，比如DFS
2. 尾递归可以用while循环来解决