算法笔记_143:构造无向图的欧拉回路(Java)

目录

1 问题描述

2 解决方案

 

---

1 问题描述

具体链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=995

---

2 解决方案

具体代码如下：

package com.liuzhen.practice;

import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    public static int MAX = 1000;
    public static int start, count;
    public static int num = 0;
    public static int[] id = new int[MAX];
    public static int[] degree = new int[MAX];  //用于计算给定图每个顶点的度
    public static boolean[] used = new boolean[MAX];   //用于判断图中相应边是否被遍历
    public static String[] path = new String[MAX];
    public static ArrayList<String> result = new ArrayList<String>();                      

    static class edge {
        public int a;  //边的起点
        public int b;  //边的终点
        public int num;  //边的编号

        public edge(int a, int b, int num) {
            this.a = a;
            this.b = b;
            this.num = num;
        }

        public String getAB() {
            return a + " "+ b;
        }
    }
    //寻找顶点a的根节点
    public int find(int[] id, int a) {
        int root = a;
        while(id[root] >= 0) {
            root = id[root];
        }
        int i;
        int k = a;
        while(k != root) {
            i = id[k];
            id[k] = root;
            k = i;
        }
        return root;
    }
    //合并顶点a和顶点b所在的树
    public void union(int[] id, int a, int b) {
        int rootA = find(id, a);
        int rootB = find(id, b);
        if(rootA == rootB)
            return;
        int rootNum = id[rootA] + id[rootB];
        if(id[rootA] < id[rootB]) {
            id[rootB] = rootA;
            id[rootA] = rootNum;
        } else{
            id[rootA] = rootB;
            id[rootB] = rootNum;
        }
        return;
    }

    public void init() {
        count = 0;
        for(int i = 0;i < 51;i++) {
            id[i] = -1;   //初始化所有顶点所在树的根节点编号为-1
            degree[i] = 0;
        }
        for(int i = 0;i < MAX;i++) {
            used[i] = false;
            path[i] = "";
        }
        return;
    }

    public boolean judge(ArrayList<edge>[] map) {
        int root = find(id, start);
        for(int i = 0;i < map.length;i++) {
            for(int j = 0;j < map[i].size();j++) {
                if(root != find(id, map[i].get(j).b))
                    return false;
            }
        }
        for(int i = 0;i < degree.length;i++) {
            if(degree[i] % 2 != 0)
                return false;
        }
        return true;
    }

    public void dfs(ArrayList<edge>[] map, int start) {
        for(int i = 0;i < map[start].size();i++) {
            if(!used[map[start].get(i).num]) {
                used[map[start].get(i).num] = true;
                path[count++] = map[start].get(i).getAB();
                dfs(map, map[start].get(i).b);
            }
        }
    }

    public static void main(String[] args) {
        Main test = new Main();
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();       //总共要输入的图的数目
        while(t > 0) {
            t--;
            @SuppressWarnings("unchecked")
            ArrayList<edge>[] map = new ArrayList[51];
            for(int i = 0;i < 51;i++)
                map[i] = new ArrayList<edge>();
            int k = in.nextInt();   //一次输入图的边数目
            test.init();
            for(int i = 0;i < k;i++) {
                int a = in.nextInt();
                int b = in.nextInt();
                map[a].add(new edge(a, b, num));
                map[b].add(new edge(b, a, num++));
                degree[a]++;
                degree[b]++;
                test.union(id, a, b);
                start = a;
            }
            String temp = "";
            if(test.judge(map)) {
                test.dfs(map, start);
                for(int i = 0;i < k;i++) {
                    temp = temp + path[i] + "\n";
                }
            } else {
                temp = "some beads may be lost";
            }
            result.add(temp);
        }
        for(int i = 0;i < result.size();i++) {
            System.out.println("Case #"+(i+1));
            System.out.println(result.get(i)+"\n");
        }
    }
}

运行结果：

2
5
1 2
2 3
3 4
4 5
5 6
5
2 1
2 2
3 4
3 1
2 4
Case #1
some beads may be lost

Case #2
2 1
1 3
3 4
4 2
2 2

参考资料：

1.欧拉回路