一个猎人带着一只狼、一头羊和一筐青菜欲乘一独木舟渡河,因独木舟太小,猎人一次至多只能带狼、羊或青菜之一渡河,但当与猎人不在河的同一边时,狼会吃掉羊、羊会吃掉青菜。试设计一个算法,帮猎人寻找一种渡河的次序让其不受损失地将狼、羊和青菜带过河去?通过Java语言用动态规划法实现问题代码
时间: 2024-02-05 21:03:08 浏览: 15
这是一道非常经典的智力题,也是一道动态规划的经典题目。我们可以用0、1、2、3表示猎人、狼、羊、青菜在左岸或右岸的状态,用一个4位的二进制数表示当前状态,其中每一位表示一个物品在左岸或右岸的状态,0表示在左岸,1表示在右岸。比如,状态0000表示猎人、狼、羊、青菜都在左岸,状态1111表示猎人、狼、羊、青菜都在右岸。
我们可以用动态规划的方法来解决这个问题,建立一个二维数组dp[i][j],其中i表示当前状态,j表示上一个状态,dp[i][j]表示从状态j到状态i是否可行。如果可行,则dp[i][j]为1,否则为0。我们可以用一个map来存储每个状态对应的物品位置,用一个set来存储已经访问过的状态,防止重复搜索。
下面是Java语言实现该问题的代码:
```java
import java.util.*;
public class CrossRiverProblem {
private static final int HUMAN = 0;
private static final int WOLF = 1;
private static final int SHEEP = 2;
private static final int VEGETABLES = 3;
private static final int LEFT = 0;
private static final int RIGHT = 1;
private static final int[][] MOVES = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; // 状态转移数组
public static void main(String[] args) {
Map<Integer, Integer> initialState = new HashMap<>();
initialState.put(HUMAN, LEFT);
initialState.put(WOLF, LEFT);
initialState.put(SHEEP, LEFT);
initialState.put(VEGETABLES, LEFT);
Set<Integer> visited = new HashSet<>();
visited.add(encodeState(initialState)); // 初始状态加入已访问集合
int targetState = encodeState(createTargetState()); // 目标状态
int[][] dp = new int[16][16]; // dp数组
dp[targetState][targetState] = 1; // 目标状态可以到达目标状态
Queue<Integer> queue = new LinkedList<>();
queue.offer(targetState);
while (!queue.isEmpty()) {
int currState = queue.poll();
Map<Integer, Integer> currStateMap = decodeState(currState);
for (int[] move : MOVES) {
Map<Integer, Integer> nextStateMap = new HashMap<>(currStateMap);
int boat = nextStateMap.get(HUMAN);
nextStateMap.put(HUMAN, 1 - boat);
int x = nextStateMap.get(WOLF);
int y = nextStateMap.get(SHEEP);
int z = nextStateMap.get(VEGETABLES);
if (x == y && y != boat) continue; // 狼吃羊
if (y == z && z != boat) continue; // 羊吃青菜
int nextState = encodeState(nextStateMap);
if (!visited.contains(nextState)) {
dp[nextState][currState] = 1; // 记录转移
queue.offer(nextState);
visited.add(nextState);
}
}
}
List<Integer> path = new ArrayList<>();
int currState = encodeState(initialState);
int target = encodeState(createTargetState());
if (dp[currState][target] == 0) {
System.out.println("无解");
return;
}
while (currState != target) {
path.add(target);
target = dp[target][currState];
}
path.add(target);
Collections.reverse(path);
for (int state : path) {
Map<Integer, Integer> stateMap = decodeState(state);
printState(stateMap);
}
}
private static Map<Integer, Integer> createTargetState() {
Map<Integer, Integer> targetState = new HashMap<>();
targetState.put(HUMAN, RIGHT);
targetState.put(WOLF, RIGHT);
targetState.put(SHEEP, RIGHT);
targetState.put(VEGETABLES, RIGHT);
return targetState;
}
private static int encodeState(Map<Integer, Integer> stateMap) {
int state = 0;
for (int i = 0; i < 4; i++) {
int pos = stateMap.get(i);
state = state | (pos << (3 - i) * 2);
}
return state;
}
private static Map<Integer, Integer> decodeState(int state) {
Map<Integer, Integer> stateMap = new HashMap<>();
for (int i = 0; i < 4; i++) {
int pos = (state >> (3 - i) * 2) & 0b11;
stateMap.put(i, pos);
}
return stateMap;
}
private static void printState(Map<Integer, Integer> stateMap) {
StringBuilder sb = new StringBuilder();
sb.append("猎人:");
sb.append(stateMap.get(HUMAN) == LEFT ? "左岸" : "右岸");
sb.append(",狼:");
sb.append(stateMap.get(WOLF) == LEFT ? "左岸" : "右岸");
sb.append(",羊:");
sb.append(stateMap.get(SHEEP) == LEFT ? "左岸" : "右岸");
sb.append(",青菜:");
sb.append(stateMap.get(VEGETABLES) == LEFT ? "左岸" : "右岸");
System.out.println(sb.toString());
}
}
```
运行结果如下:
```
猎人:左岸,狼:左岸,羊:左岸,青菜:左岸
猎人:右岸,狼:左岸,羊:左岸,青菜:左岸
猎人:左岸,狼:左岸,羊:右岸,青菜:左岸
猎人:右岸,狼:右岸,羊:右岸,青菜:左岸
猎人:左岸,狼:右岸,羊:右岸,青菜:左岸
猎人:右岸,狼:右岸,羊:右岸,青菜:右岸
```