圖論系列（四）——圖的深度優先遍歷的應用2

1. 圖的環檢測

無向圖有環：

1. 當前點的鄰接節點已經被訪問過
2. 被訪問過的鄰接節點不是當前節點的上個訪問

import java.util.ArrayList;
import java.util.Collections;

public class CycleDetection {

    private Graph G;
    private boolean[] visited;
    private boolean hasCycle = false;

    CycleDetection(Graph G) {
        this.G = G;
        visited = new boolean[G.V()];

        //解決非連通圖的遍歷問題
        for (int v = 0; v < G.V(); v++) {
            if(!visited[v])
                if(dfs(v, v)){
                    hasCycle = true;
                    break;
                }
        }
        //dfs(s, s);
    }

    //當前節點v出發是否有環
    private boolean dfs(int v, int parent) {
        visited[v] = true;

        for (int w : G.adj(v)) {
            if (!visited[w])
                if(dfs(w, v))
                    return true;
            else if(w != parent)
                return true;
        }

        return false;
    }

    public boolean hasCycle(){
        return hasCycle;
    }

    public static void main(String[] args) {

        Graph g = new Graph("g_4.txt");
        CycleDetection cycleDetection = new CycleDetection(g);
        System.out.println(cycleDetection.hasCycle());
    }
}

• 判斷一張圖是否是一棵樹

1. 圖中沒環
2. 圖聯通

2. 二分圖檢測

對於非聯通圖，只要每個連通分量是一個二分圖，整體就一定是一個二分圖

import java.util.ArrayList;

public class BipartitionDetection {

    private Graph G;

    private boolean[] visited;
    private int[] colors;       //colors[i]：節點i的顏色:0、1
    private boolean isBipartite = true;

    public BipartitionDetection(Graph G){

        this.G = G;
        visited = new boolean[G.V()];
        colors = new int[G.V()];
        for(int i = 0; i < G.V(); i ++)
            colors[i] = -1;

        for(int v = 0; v < G.V(); v ++)
            if(!visited[v])
                if(!dfs(v, 0)){
                    isBipartite = false;
                    break;
                }
    }

    private boolean dfs(int v, int color){

        visited[v] = true;
        colors[v] = color;          //節點v染成color

        for(int w: G.adj(v))
            if(!visited[w]){
                if(!dfs(w, 1 - color)) return false;
            }
            else if(colors[w] == colors[v])     //當前節點與訪問過的鄰接節點顏色一致
                return false;
        return true;
    }

    public boolean isBipartite(){
        return isBipartite;
    }

    public static void main(String[] args){

        Graph g = new Graph("g_4_1.txt");
        BipartitionDetection bipartitionDetection = new BipartitionDetection(g);
        System.out.println(bipartitionDetection.isBipartite());
    }
}

• 注：本系列參考玩轉算法系列–圖論精講