假如你是一位Java工程师，请你写一份关于农夫过河详细的代码，使用Java语言。谢谢

好的，以下是关于农夫过河问题的Java代码实现：

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;

class State {
    int farmer; // 0 表示农夫在起点岸，1 表示农夫在目标岸
    int wolf;   // 0 表示狼在起点岸，1 表示狼在目标岸
    int goat;   // 0 表示山羊在起点岸，1 表示山羊在目标岸
    int cabbage;// 0 表示卷心菜在起点岸，1 表示卷心菜在目标岸

    boolean isLegal() {
        if (goat == cabbage && farmer != goat) { //山羊吃卷心菜
            return false;
        } else if (wolf == goat && farmer != goat) { //狼吃山羊
            return false;
        } else {
            return true;
        }
    }

    boolean isGoal() {
        return farmer == 1 && goat == 1 && wolf == 1 && cabbage == 1;
    }

    State(int farmer, int wolf, int goat, int cabbage) {
        this.farmer = farmer;
        this.wolf = wolf;
        this.goat = goat;
        this.cabbage = cabbage;
    }

    State(State other) {
        this.farmer = other.farmer;
        this.wolf = other.wolf;
        this.goat = other.goat;
        this.cabbage = other.cabbage;
    }

    public boolean equals(Object obj) {
        if (obj == null || !(obj instanceof State)) {
            return false;
        }
        State other = (State) obj;
        return farmer == other.farmer && wolf == other.wolf && goat == other.goat && cabbage == other.cabbage;
    }

    public int hashCode() {
        return farmer * 8 + wolf * 4 + goat * 2 + cabbage;
    }
}

public class FarmerWolfGoatCabbage {
    public static void main(String[] args) {
        bfs(new State(0, 0, 0, 0));
    }

    private static void bfs(State start) {
        Queue<State> queue = new LinkedList<>();
        Set<State> visited = new HashSet<>();

        queue.offer(start);
        visited.add(start);

        while (!queue.isEmpty()) {
            State curr = queue.poll();
            if (curr.isGoal()) {
                printPath(curr);
                return;
            }

            for (State next : getNextStates(curr)) {
                if (next.isLegal() && !visited.contains(next)) {
                    visited.add(next);
                    queue.offer(next);
                }
            }
        }
    }

    private static void printPath(State state) {
        StringBuilder sb = new StringBuilder();
        while (state != null) {
            sb.insert(0, state.farmer + "" + state.wolf + "" + state.goat + "" + state.cabbage + "\n");
            state = state.parent;
        }
        System.out.println(sb.toString());
    }

    private static State[] getNextStates(State curr) {
        State[] nextStates = new State[8];
        int farmer = curr.farmer == 0 ? 1 : 0;

        //农夫过河
        for (int i = 0; i < 4; i++) {
            if (i == 0) { //农夫自己过河
                nextStates[0] = new State(farmer, curr.wolf, curr.goat, curr.cabbage);
                nextStates[0].parent = curr;
            } else if (i == 1 && curr.wolf == curr.farmer) { //狼和农夫一起过河
                nextStates[1] = new State(farmer, farmer, curr.goat, curr.cabbage);
                nextStates[1].parent = curr;
            } else if (i == 2 && curr.goat == curr.farmer) { //羊和农夫一起过河
                nextStates[2] = new State(farmer, curr.wolf, farmer, curr.cabbage);
                nextStates[2].parent = curr;
            } else if (i == 3 && curr.cabbage == curr.farmer) { //卷心菜和农夫一起过河
                nextStates[3] = new State(farmer, curr.wolf, curr.goat, farmer);
                nextStates[3].parent = curr;
            }
        }

        return nextStates;
    }
}

这里使用了广度优先搜索算法求解答案，将当前状态视为节点，一次过河视为一次转移。getNextStates 函数负责根据当前状态生成所有合法的下一个状态，isLegal 函数判断当前状态是否合法，isGoal 函数判断当前状态是否是目标状态。最终的答案通过回溯得到。