有n个牧师和n个野人准备渡河，但只有一条能容纳c个人的小船，为了防止野人侵犯牧师，要求无论在何处，牧师的人数不得少于野人的人数(除非牧师人数为0)，且假定野人与牧师都会划船，试设计一个算法，确定他们能否渡过河去，若能，则给出小船来回次数最少的最佳方案。

这是一个经典的河岸问题，可以使用深度优先搜索或广度优先搜索来解决。以下是一种可能的算法：

1. 定义状态：用一个三元组 (M, C, B) 表示当前状态，其中 M 表示左岸牧师的数量，C 表示左岸野人的数量，B 表示船的位置，B=0 表示船在左岸，B=1 表示船在右岸。

2. 定义操作：每次可以将船上最多 c 个人从一个岸移动到另一个岸，但必须满足以下条件：左岸和右岸的牧师数量都不得少于野人数量（除非牧师数量为0），且船上的人数不能超过 c。

3. 定义目标：将所有牧师和野人都移动到右岸。

4. 使用深度优先搜索或广度优先搜索来搜索所有可能的状态，直到找到一种方案可以达到目标状态。

5. 在搜索过程中，记录每个状态的父状态，以便在找到目标状态后回溯找到最佳方案。

6. 最佳方案是指船来回次数最少的方案。可以通过记录每个状态的深度来计算船来回的次数。

7. 如果找不到任何一种方案可以达到目标状态，则说明无法渡过河去。

注意：在搜索过程中，需要避免重复访问已经访问过的状态，否则会导致无限循环。可以使用哈希表或其他数据结构来记录已经访问过的状态。

相关问题

用c++代码实现假如有n个牧师和n个野人准备渡河，但只有一条能容纳c个人的小船，为了防止野人侵犯牧师，要求无论在何处，牧师的人数不得少于野人的人数（除非牧师人数为0），且假定两种人都会划船，试设计一个算法，确定它们能否渡过河去，若能，则给出一只小船来回次数最少的最佳方案。

以下是一个基于DFS的解法，代码中使用了vector来存储船上的人员，方便操作和输出。

#include <iostream>
#include <vector>

using namespace std;

const int MAXN = 10;

int n, c; // n个牧师和n个野人，小船容纳c个人
bool visited[MAXN + 1][MAXN + 1];
vector<int> left_bank; // 左岸人员
vector<int> right_bank; // 右岸人员

bool check(vector<int>& bank) { // 检查当前状态是否合法
    int p = 0, m = 0;
    for (int i = 0; i < bank.size(); i++) {
        if (bank[i] == 1) p++;
        else m++;
    }
    if (p < m) return false;
    if (n - p < n - m) return false;
    return true;
}

void dfs(int step) {
    if (right_bank.size() == n) { // 右岸已经全部通过
        cout << "Solution:\n";
        for (int i = 0; i < right_bank.size(); i++) {
            cout << right_bank[i];
        }
        cout << endl;
        return;
    }
    if (step % 2 == 0) { // 船在左岸
        for (int i = 0; i < left_bank.size(); i++) {
            for (int j = 0; j < left_bank.size(); j++) {
                if (i == j) continue; // 不能选择同一个人
                vector<int> temp = left_bank;
                vector<int> boat;
                boat.push_back(temp[i]);
                boat.push_back(temp[j]);
                for (int k = 0; k < boat.size(); k++) {
                    temp.erase(find(temp.begin(), temp.end(), boat[k]));
                }
                if (check(temp) && !visited[temp.size()][boat.size()]) { // 检查合法性
                    visited[temp.size()][boat.size()] = true;
                    right_bank.insert(right_bank.end(), boat.begin(), boat.end());
                    dfs(step + 1);
                    visited[temp.size()][boat.size()] = false;
                    for (int k = 0; k < boat.size(); k++) {
                        right_bank.pop_back();
                    }
                }
            }
        }
    } else { // 船在右岸
        for (int i = 0; i < right_bank.size(); i++) {
            for (int j = 0; j < right_bank.size(); j++) {
                if (i == j) continue;
                vector<int> temp = right_bank;
                vector<int> boat;
                boat.push_back(temp[i]);
                boat.push_back(temp[j]);
                for (int k = 0; k < boat.size(); k++) {
                    temp.erase(find(temp.begin(), temp.end(), boat[k]));
                }
                if (check(temp) && !visited[temp.size()][c - boat.size()]) {
                    visited[temp.size()][c - boat.size()] = true;
                    left_bank.insert(left_bank.end(), boat.begin(), boat.end());
                    dfs(step + 1);
                    visited[temp.size()][c - boat.size()] = false;
                    for (int k = 0; k < boat.size(); k++) {
                        left_bank.pop_back();
                    }
                }
            }
        }
    }
}

int main() {
    cout << "Please input the number of people and the capacity of the boat:" << endl;
    cin >> n >> c;
    for (int i = 0; i < n; i++) {
        left_bank.push_back(1);
        left_bank.push_back(0);
    }
    dfs(0);
    return 0;
}

代码中使用的visited数组用来记录当前状态是否已经被访问过，避免重复搜索，同时也可以剪枝。

用Java编写过河问题：问题转变为：假如有n个牧师和n个野人准备渡河，但只有一条能容纳c个人的小船，为了防止野人侵犯牧师，要求无论在何处，牧师的人数不得少于野人的人数（除非牧师人数为0），且假定两种人都会划船，试设计一个算法，确定它们能否渡过河去，若能，则给出一只小船来回次数最少的最佳方案。

以下是用Java编写的河问题的解法：

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class RiverCrossing {

    private static final int NUM_OF_PASSENGERS = 3; // 每次最多能过河的人数
    private static final int NUM_OF_PASSENGERS_TYPE = 2; // 人的种类数量（牧师和野人）
    private static final int NUM_OF_BOAT_SEATS = 2; // 小船可以坐的人数
    private static final int START_STATE = 0; // 起始状态
    private static final int END_STATE = (1 << (NUM_OF_PASSENGERS * NUM_OF_PASSENGERS_TYPE)) - 1; // 终止状态

    private static final int LEFT_BANK = 0; // 左岸
    private static final int RIGHT_BANK = 1; // 右岸

    private static final int MISSIONARY = 0; // 牧师
    private static final int CANNIBAL = 1; // 野人

    // 判断当前状态是否合法
    private static boolean isValid(int state) {
        int leftBank[] = decodeState(state, LEFT_BANK);
        int rightBank[] = decodeState(state, RIGHT_BANK);
        if (leftBank[MISSIONARY] != 0 && leftBank[MISSIONARY] < leftBank[CANNIBAL])
            return false;
        if (rightBank[MISSIONARY] != 0 && rightBank[MISSIONARY] < rightBank[CANNIBAL])
            return false;
        return true;
    }

    // 将状态转换为数组
    private static int[] decodeState(int state, int bank) {
        int[] bankArray = new int[NUM_OF_PASSENGERS_TYPE];
        for (int i = 0; i < NUM_OF_PASSENGERS; i++) {
            int passengerType = (state >> (i * NUM_OF_PASSENGERS_TYPE + bank * NUM_OF_PASSENGERS_TYPE * NUM_OF_PASSENGERS))
                    & (NUM_OF_PASSENGERS_TYPE - 1);
            bankArray[passengerType]++;
        }
        return bankArray;
    }

    // 将数组转换为状态
    private static int encodeState(int[] leftBank, int[] rightBank) {
        int state = 0;
        for (int i = 0; i < NUM_OF_PASSENGERS; i++) {
            state |= (leftBank[MISSIONARY] > i ? MISSIONARY : CANNIBAL)
                    << (i * NUM_OF_PASSENGERS_TYPE + LEFT_BANK * NUM_OF_PASSENGERS_TYPE * NUM_OF_PASSENGERS);
            state |= (rightBank[MISSIONARY] > i ? MISSIONARY : CANNIBAL)
                    << (i * NUM_OF_PASSENGERS_TYPE + RIGHT_BANK * NUM_OF_PASSENGERS_TYPE * NUM_OF_PASSENGERS);
        }
        return state;
    }

    // 判断状态是否是终止状态
    private static boolean isEndState(int state) {
        return state == END_STATE;
    }

    // 判断两个状态是否相等
    private static boolean isEqual(int state1, int state2) {
        return state1 == state2;
    }

    // 判断一个状态是否已经被访问过
    private static boolean isVisited(int state, Set<Integer> visitedStates) {
        return visitedStates.contains(state);
    }

    // 获取下一个可能的状态
    private static List<Integer> getNextPossibleStates(int state) {
        List<Integer> nextPossibleStates = new ArrayList<>();
        int[] leftBank = decodeState(state, LEFT_BANK);
        int[] rightBank = decodeState(state, RIGHT_BANK);
        int boatPosition = LEFT_BANK;
        if (leftBank[NUM_OF_PASSENGERS_TYPE - 1] == NUM_OF_PASSENGERS
                && isEndState(encodeState(rightBank, leftBank))) {
            nextPossibleStates.add(encodeState(leftBank, rightBank));
            return nextPossibleStates;
        }
        if (rightBank[NUM_OF_PASSENGERS_TYPE - 1] == NUM_OF_PASSENGERS
                && isEndState(encodeState(leftBank, rightBank))) {
            nextPossibleStates.add(encodeState(leftBank, rightBank));
            return nextPossibleStates;
        }
        if (leftBank[MISSIONARY] >= 1 && boatPosition == LEFT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newLeftBank[MISSIONARY]--;
            newRightBank[MISSIONARY]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (leftBank[CANNIBAL] >= 1 && boatPosition == LEFT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newLeftBank[CANNIBAL]--;
            newRightBank[CANNIBAL]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (leftBank[MISSIONARY] >= 2 && boatPosition == LEFT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newLeftBank[MISSIONARY] -= 2;
            newRightBank[MISSIONARY] += 2;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (leftBank[CANNIBAL] >= 2 && boatPosition == LEFT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newLeftBank[CANNIBAL] -= 2;
            newRightBank[CANNIBAL] += 2;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (leftBank[MISSIONARY] >= 1 && leftBank[CANNIBAL] >= 1 && boatPosition == LEFT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newLeftBank[MISSIONARY]--;
            newLeftBank[CANNIBAL]--;
            newRightBank[MISSIONARY]++;
            newRightBank[CANNIBAL]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (rightBank[MISSIONARY] >= 1 && boatPosition == RIGHT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newRightBank[MISSIONARY]--;
            newLeftBank[MISSIONARY]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (rightBank[CANNIBAL] >= 1 && boatPosition == RIGHT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newRightBank[CANNIBAL]--;
            newLeftBank[CANNIBAL]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (rightBank[MISSIONARY] >= 2 && boatPosition == RIGHT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newRightBank[MISSIONARY] -= 2;
            newLeftBank[MISSIONARY] += 2;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (rightBank[CANNIBAL] >= 2 && boatPosition == RIGHT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newRightBank[CANNIBAL] -= 2;
            newLeftBank[CANNIBAL] += 2;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        if (rightBank[MISSIONARY] >= 1 && rightBank[CANNIBAL] >= 1 && boatPosition == RIGHT_BANK) {
            int[] newLeftBank = leftBank.clone();
            int[] newRightBank = rightBank.clone();
            newRightBank[MISSIONARY]--;
            newRightBank[CANNIBAL]--;
            newLeftBank[MISSIONARY]++;
            newLeftBank[CANNIBAL]++;
            if (isValid(encodeState(newLeftBank, newRightBank)))
                nextPossibleStates.add(encodeState(newLeftBank, newRightBank));
        }
        return nextPossibleStates;
    }

    // 搜索最短路径
    private static List<Integer> search() {
        List<Integer> shortestPath = new ArrayList<>();
        Set<Integer> visitedStates = new HashSet<>();
        List<List<Integer>> paths = new ArrayList<>();
        List<Integer> startPath = new ArrayList<>();
        startPath.add(START_STATE);
        paths.add(startPath);

        while (!paths.isEmpty()) {
            List<Integer> currentPath = paths.remove(0);
            int currentState = currentPath.get(currentPath.size() - 1);
            if (isEndState(currentState)) {
                if (shortestPath.isEmpty() || currentPath.size() < shortestPath.size())
                    shortestPath = currentPath;
                continue;
            }
            visitedStates.add(currentState);
            List<Integer> nextPossibleStates = getNextPossibleStates(currentState);
            for (int nextState : nextPossibleStates) {
                if (isVisited(nextState, visitedStates))
                    continue;
                List<Integer> nextPath = new ArrayList<>(currentPath);
                nextPath.add(nextState);
                paths.add(nextPath);
            }
        }
        return shortestPath;
    }

    // 打印结果
    private static void printResult(List<Integer> path) {
        System.out.println("Number of boat trips: " + (path.size() - 1));
        for (int i = 0; i < path.size() - 1; i++) {
            int[] leftBank = decodeState(path.get(i), LEFT_BANK);
            int[] rightBank = decodeState(path.get(i), RIGHT_BANK);
            int[] nextLeftBank = decodeState(path.get(i + 1), LEFT_BANK);
            int[] nextRightBank = decodeState(path.get(i + 1), RIGHT_BANK);

            System.out.println("From left bank to right bank: ");
            System.out.println("Missionaries: " + (leftBank[MISSIONARY] - nextLeftBank[MISSIONARY]));
            System.out.println("Cannibals: " + (leftBank[CANNIBAL] - nextLeftBank[CANNIBAL]));

            System.out.println("From right bank to left bank: ");
            System.out.println("Missionaries: " + (rightBank[MISSIONARY] - nextRightBank[MISSIONARY]));
            System.out.println("Cannibals: " + (rightBank[CANNIBAL] - nextRightBank[CANNIBAL]));
        }
    }

    public static void main(String[] args) {
        List<Integer> path = search();
        printResult(path);
    }
}

该程序将所有可能的状态表示为一个二进制数，并使用两个数组来表示左岸和右岸上的每种人的数量。搜索算法使用BFS，并利用isValid，encodeState和decodeState函数来处理状态。getNextPossibleStates函数返回下一组可能的状态。程序搜索最短路径，然后打印出每次小船的行动。