JAVA野人_Java实现传教士与野人过河问题
时间: 2023-12-09 07:03:10 浏览: 48
传教士与野人过河问题可以使用深度优先搜索算法(DFS)来实现。以下是Java代码实现:
```java
import java.util.ArrayList;
public class MissionariesAndCannibals {
private static final int numMissionaries = 3;
private static final int numCannibals = 3;
private static final int boatCapacity = 2;
private static final int startingBank = 0;
private static final int oppositeBank = 1;
private static final int MAX_DEPTH = 100;
private static boolean[][][] visitedStates = new boolean[numMissionaries + 1][numCannibals + 1][2];
private static ArrayList<String> solution = new ArrayList<>();
public static void main(String[] args) {
dfs(startingBank, numMissionaries, numCannibals, "");
printSolution();
}
private static void dfs(int bank, int numMissionariesLeft, int numCannibalsLeft, String path) {
if (numMissionariesLeft == 0 && numCannibalsLeft == 0) {
solution.add(path);
return;
}
if (path.length() >= MAX_DEPTH) {
return;
}
visitedStates[numMissionariesLeft][numCannibalsLeft][bank] = true;
if (bank == startingBank) {
for (int i = 0; i <= Math.min(numMissionariesLeft, boatCapacity); i++) {
for (int j = 0; j <= Math.min(numCannibalsLeft, boatCapacity); j++) {
if (i + j == 0 || i + j > boatCapacity) {
continue;
}
int newMissionariesLeft = numMissionariesLeft - i;
int newCannibalsLeft = numCannibalsLeft - j;
if (newMissionariesLeft < newCannibalsLeft && newMissionariesLeft > 0) {
continue;
}
String nextPath = path + "(" + i + "," + j + ") ";
dfs(oppositeBank, newMissionariesLeft, newCannibalsLeft, nextPath);
}
}
} else {
for (int i = 0; i <= Math.min(numMissionaries - numMissionariesLeft, boatCapacity); i++) {
for (int j = 0; j <= Math.min(numCannibals - numCannibalsLeft, boatCapacity); j++) {
if (i + j == 0 || i + j > boatCapacity) {
continue;
}
int newMissionariesLeft = numMissionariesLeft + i;
int newCannibalsLeft = numCannibalsLeft + j;
if (newMissionariesLeft < newCannibalsLeft && newMissionariesLeft > 0) {
continue;
}
String nextPath = path + "(" + i + "," + j + ") ";
dfs(startingBank, newMissionariesLeft, newCannibalsLeft, nextPath);
}
}
}
}
private static void printSolution() {
for (String step : solution) {
System.out.println(step);
}
}
}
```
在这个实现中,我们使用 `dfs` 方法来递归地遍历所有可能的状态。我们使用一个 `visitedStates` 数组来避免重复访问同一个状态,以避免死循环。在每个递归层次,我们检查当前状态是否为解决方案,如果是,则将路径添加到 `solution` 列表中。如果路径长度超过 `MAX_DEPTH`,则返回。
在每个状态中,我们遍历所有可能的移动选择,将每个选择作为一个新的状态传递给 `dfs`。我们只考虑有效的移动,即移动不得超过船的容量、传教士数量不能少于野人数量、传教士数量不能为负数等等。
最后,我们打印出所有解决方案的路径。
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