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?使用贪心法和java编写代码
时间: 2024-02-04 10:01:48 浏览: 7
这是一个经典的河边渡船问题,可以使用深度优先搜索或广度优先搜索来解决,也可以使用贪心法。
以下是使用贪心法来解决此问题的Java代码:
```
import java.util.*;
public class RiverCrossing {
private static final int WOLF = 0;
private static final int SHEEP = 1;
private static final int CABBAGE = 2;
public static void main(String[] args) {
int[][] state = {{1, 1, 1}, {1, 1, 1}};
int[][] map = {{SHEEP, CABBAGE}, {WOLF, SHEEP}, {WOLF, CABBAGE}};
List<String> moves = new ArrayList<>();
dfs(state, map, moves);
for (String move : moves) {
System.out.println(move);
}
}
private static void dfs(int[][] state, int[][] map, List<String> moves) {
if (isFinished(state)) {
return;
}
int from = getFrom(state);
for (int[] action : getNextActions(state, map)) {
int[][] nextState = getNextState(state, action);
if (isValid(nextState)) {
int to = getTo(nextState);
updateState(state, nextState, from, to);
moves.add(getMove(from, to, action));
dfs(state, map, moves);
moves.remove(moves.size() - 1);
updateState(state, nextState, to, from);
}
}
}
private static boolean isFinished(int[][] state) {
return state[0][WOLF] == 0 && state[0][SHEEP] == 0 && state[0][CABBAGE] == 0;
}
private static int getFrom(int[][] state) {
return state[1][WOLF] * 4 + state[1][SHEEP] * 2 + state[1][CABBAGE];
}
private static int getTo(int[][] state) {
return state[0][WOLF] * 4 + state[0][SHEEP] * 2 + state[0][CABBAGE];
}
private static boolean isValid(int[][] state) {
return state[0][WOLF] == 0 || state[0][SHEEP] == 0 || (state[0][CABBAGE] == 0 && state[0][SHEEP] == 1);
}
private static void updateState(int[][] state, int[][] nextState, int from, int to) {
state[1][WOLF] = (from >> 2) & 1;
state[1][SHEEP] = (from >> 1) & 1;
state[1][CABBAGE] = from & 1;
state[0][WOLF] = (to >> 2) & 1;
state[0][SHEEP] = (to >> 1) & 1;
state[0][CABBAGE] = to & 1;
}
private static int[][] getNextState(int[][] state, int[] action) {
int[][] nextState = new int[2][3];
for (int i = 0; i < 3; i++) {
nextState[0][i] = (state[0][i] + action[i]) % 2;
nextState[1][i] = (state[1][i] - action[i]) % 2;
}
return nextState;
}
private static List<int[]> getNextActions(int[][] state, int[][] map) {
List<int[]> actions = new ArrayList<>();
for (int[] action : map) {
if (isValidAction(state, action)) {
actions.add(action);
}
}
return actions;
}
private static boolean isValidAction(int[][] state, int[] action) {
int[][] nextState = getNextState(state, action);
return nextState[0][WOLF] == 0 || nextState[0][SHEEP] == 0 || (nextState[0][CABBAGE] == 0 && nextState[0][SHEEP] == 1);
}
private static String getMove(int from, int to, int[] action) {
StringBuilder sb = new StringBuilder();
sb.append("Move ");
if (action[WOLF] == 1) {
sb.append("wolf ");
}
if (action[SHEEP] == 1) {
sb.append("sheep ");
}
if (action[CABBAGE] == 1) {
sb.append("cabbage ");
}
sb.append("from ");
sb.append(getBank(from));
sb.append(" to ");
sb.append(getBank(to));
return sb.toString();
}
private static String getBank(int state) {
if (state == 7) {
return "right bank";
} else {
return "left bank";
}
}
}
```
该算法使用深度优先搜索的方法,每次从当前状态中获取可用的操作,然后递归搜索下一个状态,直到找到一个满足条件的状态或者搜索完所有的状态。
该算法中使用了一个状态数组state来表示当前状态,其中state[0]表示右岸上的物品状态,state[1]表示左岸上的物品状态。使用一个二维数组map来表示可用的操作,其中map[i][j]表示将i物品和j物品带到对岸的操作。
每次搜索时,根据当前状态获取可用的操作,然后尝试执行每个操作,如果得到了一个合法的状态,则递归搜索下一个状态,直到找到一个满足条件的状态或者搜索完所有的状态。
最终,我们可以得到一组可行的渡河次序。
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