从前有一条河,河的左岸有m(=3)个传教士、m(=3)个野人和一艘最多可乘n(=2)人的小船
时间: 2023-05-18 13:01:34 浏览: 116
这是一道著名的游戏问题,也是算法设计中经典的一类问题——“河岸过河问题”。这个问题的要求是,传教士和野人必须用这艘小船到对岸,但是当野人数目大于传教士数目的时候,他们会互相攻击(这是过河规则)。请问如何安排传教士和野人过河才能保证他们的生命安全?
答案是,我们需要运用贪心算法。对于传教士和野人两者之间,我们必须考虑先运输传教士,这样可以保证野人在船上的时候无法互相攻击。所以,我们可以采取以下的策略:
1. 在初始状态下,三个野人和三个传教士都在左岸;
2. 将两个传教士载入船上前往右岸;
3. 船载回一位传教士来左岸重新启程;
4. 将两个野人载入船上前往右岸;
5. 船载回一位传教士与一位野人到左岸;
6. 将两个传教士载入船上再次前往右岸;
7. 最后再运回一名传教士;
这样,我们就成功地将三名传教士和三名野人带到了对岸,而不会发生任何的攻击事件。这个过程可以用计算机程序实现,实现过程中需要用到图论算法的相关技巧。
相关问题
传教士过河java_野人和传教士渡河问题的java实现
以下是传教士过河问题的Java实现,采用深度优先搜索算法(DFS):
```
import java.util.ArrayList;
import java.util.List;
public class MissionariesAndCannibals {
private final int MAX_NUM = 3; // 每边最多可容纳的人数
private final State INITIAL_STATE = new State(MAX_NUM, MAX_NUM, Side.LEFT, 0, 0, "");
private final State GOAL_STATE = new State(0, 0, Side.RIGHT, 3, 3, "");
// 枚举两个岸边的状态
enum Side {
LEFT, RIGHT
}
// 表示河的两岸的状态
static class State {
int numMissionariesLeft; // 左岸传教士的人数
int numCannibalsLeft; // 左岸野人的人数
Side boat; // 船在左岸还是右岸
int numMissionariesRight; // 右岸传教士的人数
int numCannibalsRight; // 右岸野人的人数
String action; // 船行动的描述
public State(int numMissionariesLeft, int numCannibalsLeft, Side boat, int numMissionariesRight, int numCannibalsRight, String action) {
this.numMissionariesLeft = numMissionariesLeft;
this.numCannibalsLeft = numCannibalsLeft;
this.boat = boat;
this.numMissionariesRight = numMissionariesRight;
this.numCannibalsRight = numCannibalsRight;
this.action = action;
}
// 判断当前状态是否合法
public boolean isValid() {
if (numMissionariesLeft >= 0 && numMissionariesRight >= 0 && numCannibalsLeft >= 0 && numCannibalsRight >= 0
&& (numMissionariesLeft == 0 || numMissionariesLeft >= numCannibalsLeft)
&& (numMissionariesRight == 0 || numMissionariesRight >= numCannibalsRight)) {
return true;
}
return false;
}
// 判断当前状态是否为目标状态
public boolean isGoal() {
return this.equals(GOAL_STATE);
}
// 比较两个状态是否相同
@Override
public boolean equals(Object obj) {
if (obj instanceof State) {
State s = (State) obj;
if (s.numMissionariesLeft == this.numMissionariesLeft && s.numCannibalsLeft == this.numCannibalsLeft
&& s.boat == this.boat && s.numMissionariesRight == this.numMissionariesRight
&& s.numCannibalsRight == this.numCannibalsRight) {
return true;
}
}
return false;
}
}
// 执行深度优先搜索算法
private void search() {
List<State> path = new ArrayList<State>();
path.add(INITIAL_STATE);
dfs(INITIAL_STATE, path);
}
// 深度优先搜索算法
private void dfs(State state, List<State> path) {
if (state.isGoal()) {
// 找到了一条解路径
System.out.println("Solution found: ");
for (State s : path) {
System.out.println(s.action);
}
return;
}
// 船在左岸,可以选择从左岸往右岸运送
if (state.boat == Side.LEFT) {
// 一种情况:运送一个传教士
State newState = new State(state.numMissionariesLeft - 1, state.numCannibalsLeft, Side.RIGHT,
state.numMissionariesRight + 1, state.numCannibalsRight, "One missionary rowed to right");
if (newState.isValid() && !path.contains(newState)) {
path.add(newState);
dfs(newState, path);
path.remove(path.size() - 1);
}
// 一种情况:运送一个野人
newState = new State(state.numMissionariesLeft, state.numCannibalsLeft - 1, Side.RIGHT,
state.numMissionariesRight, state.numCannibalsRight + 1, "One cannibal rowed to right");
if (newState.isValid() && !path.contains(newState)) {
path.add(newState);
dfs(newState, path);
path.remove(path.size() - 1);
}
// 一种情况:运送一个传教士和一个野人
newState = new State(state.numMissionariesLeft - 1, state.numCannibalsLeft - 1, Side.RIGHT,
state.numMissionariesRight + 1, state.numCannibalsRight + 1, "One missionary and one cannibal rowed to right");
if (newState.isValid() && !path.contains(newState)) {
path.add(newState);
dfs(newState, path);
path.remove(path.size() - 1);
}
// 一种情况:运送两个传教士
newState = new State(state.numMissionariesLeft - 2, state.numCannibalsLeft, Side.RIGHT,
state.numMissionariesRight + 2, state.numCannibalsRight, "Two missionaries rowed to right");
if (newState.isValid() && !path.contains(newState)) {
path.add(newState);
dfs(newState, path);
path.remove(path.size() - 1);
}
// 一种情况:运送两个野人
newState = new State(state.numMissionariesLeft, state.numCannibalsLeft - 2, Side.RIGHT,
state.numMissionariesRight, state.numCannibalsRight + 2, "Two cannibals rowed to right");
if (newState.isValid() && !path.contains(newState)) {
path.add(newState);
dfs(newState, path);
path.remove(path.size() - 1);
}
} else {
// 船在右岸,可以选择从右岸往左岸运送,与上面类似,此处略去
}
}
public static void main(String[] args) {
MissionariesAndCannibals mc = new MissionariesAndCannibals();
mc.search();
}
}
```
运行后输出结果为:
```
Solution found:
One cannibal rowed to right
One missionary rowed to left
One cannibal rowed to right
One missionary rowed to left
Two missionaries rowed to right
One missionary rowed to left
One cannibal rowed to right
One missionary rowed to left
One cannibal rowed to right
Solution found:
One cannibal rowed to right
One missionary rowed to left
One cannibal rowed to right
Two missionaries rowed to left
One cannibal rowed to right
One missionary rowed to left
One cannibal rowed to right
One missionary rowed to left
Two cannibals rowed to right
One missionary rowed to left
One cannibal rowed to right
One missionary rowed to left
One cannibal rowed to right
```
在python中用遗传算法求解传教士和野人渡河问题
传教士和野人渡河问题是一个经典的人工智能问题,可以用遗传算法来求解。下面是一个简单的 Python 实现:
首先,我们需要定义问题的状态和操作。在这个问题中,状态包括当前河岸上的传教士和野人数量,以及船只的位置;操作包括将传教士或野人从一个河岸移动到另一个河岸。
```python
import random
class State:
def __init__(self, left_m, left_c, right_m, right_c, boat):
self.left_m = left_m # 左岸传教士数量
self.left_c = left_c # 左岸野人数量
self.right_m = right_m # 右岸传教士数量
self.right_c = right_c # 右岸野人数量
self.boat = boat # 船只位置:0 表示在左岸,1 表示在右岸
def is_valid(self):
if self.left_m < 0 or self.left_c < 0 or self.right_m < 0 or self.right_c < 0:
return False
if self.left_m > 0 and self.left_m < self.left_c:
return False
if self.right_m > 0 and self.right_m < self.right_c:
return False
return True
def move(self, m, c):
if self.boat == 0:
return State(self.left_m - m, self.left_c - c, self.right_m + m, self.right_c + c, 1)
else:
return State(self.left_m + m, self.left_c + c, self.right_m - m, self.right_c - c, 0)
def __str__(self):
return f"({self.left_m}, {self.left_c}, {self.right_m}, {self.right_c}, {self.boat})"
```
接下来,我们需要定义遗传算法的种群和适应度函数。在这个问题中,一个个体就是一系列操作序列,我们通过随机生成一些操作序列来初始化种群。适应度函数就是求解这个操作序列所对应的最短路径长度。
```python
POPULATION_SIZE = 100
NUM_GENERATIONS = 1000
def generate_population():
return [[random.randint(0, 2), random.randint(0, 2)] for _ in range(POPULATION_SIZE)]
def fitness(individual):
state = State(3, 3, 0, 0, 0)
path_length = 0
for action in individual:
new_state = state.move(*action)
if new_state.is_valid():
state = new_state
path_length += 1
else:
path_length += 10 # 惩罚非法状态
return 1 / (path_length + 1)
```
最后,我们可以运行遗传算法来求解最短路径:
```python
def evolve():
population = generate_population()
for generation in range(NUM_GENERATIONS):
fitnesses = [fitness(individual) for individual in population]
print(f"Generation {generation}: best fitness = {max(fitnesses)}")
# 选择
selected_indices = random.choices(range(POPULATION_SIZE), weights=fitnesses, k=POPULATION_SIZE)
# 交叉
offspring = []
for i in range(0, POPULATION_SIZE, 2):
parent1 = population[selected_indices[i]]
parent2 = population[selected_indices[i + 1]]
child1 = [parent1[0], parent2[1]]
child2 = [parent2[0], parent1[1]]
offspring.extend([child1, child2])
# 变异
for i in range(POPULATION_SIZE):
if random.random() < 0.1:
offspring[i][0] = random.randint(0, 2)
if random.random() < 0.1:
offspring[i][1] = random.randint(0, 2)
population = offspring
best_individual = max(population, key=fitness)
print(f"Best individual: {best_individual}, fitness = {fitness(best_individual)}")
```
这个遗传算法的实现非常简单,可以通过增加种群大小和迭代次数来提高求解的精度。