请帮我用c语言写一下人狼羊菜过河问题的代码
时间: 2023-12-03 12:43:59 浏览: 114
农夫狼羊菜的c语言实现代码
好的,以下是人狼羊菜过河问题的C语言代码实现:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// 定义过河岸边的角色标识符常量
#define FARMER 0
#define WOLF 1
#define GOAT 2
#define CABBAGE 3
// 定义角色过河状态标识符常量
#define LEFT 0
#define RIGHT 1
// 定义过河状态结构体
struct State {
int boat; // 船的位置,0表示在左岸,1表示在右岸
int farmer; // 农夫的位置,0表示在左岸,1表示在右岸
int wolf; // 狼的位置,0表示在左岸,1表示在右岸
int goat; // 羊的位置,0表示在左岸,1表示在右岸
int cabbage; // 白菜的位置,0表示在左岸,1表示在右岸
};
// 判断当前状态是否有效,即是否存在矛盾状态(狼吃羊、羊吃白菜)
int isValidState(struct State state) {
if (state.farmer == LEFT && state.wolf == LEFT && state.goat == LEFT) {
return 0; // 狼吃羊
}
if (state.farmer == LEFT && state.goat == LEFT && state.cabbage == LEFT) {
return 0; // 羊吃白菜
}
if (state.farmer == RIGHT && state.wolf == RIGHT && state.goat == RIGHT) {
return 0; // 狼吃羊
}
if (state.farmer == RIGHT && state.goat == RIGHT && state.cabbage == RIGHT) {
return 0; // 羊吃白菜
}
return 1;
}
// 判断当前状态是否为最终状态(所有角色都在右岸)
int isFinalState(struct State state) {
return (state.farmer == RIGHT && state.wolf == RIGHT && state.goat == RIGHT && state.cabbage == RIGHT);
}
// 复制状态
struct State copyState(struct State state) {
struct State newState;
newState.boat = state.boat;
newState.farmer = state.farmer;
newState.wolf = state.wolf;
newState.goat = state.goat;
newState.cabbage = state.cabbage;
return newState;
}
// 移动狼
struct State moveWolf(struct State state) {
if (state.boat == LEFT) {
state.wolf = RIGHT;
} else {
state.wolf = LEFT;
}
state.farmer = !state.farmer;
state.boat = !state.boat;
return state;
}
// 移动羊
struct State moveGoat(struct State state) {
if (state.boat == LEFT) {
state.goat = RIGHT;
} else {
state.goat = LEFT;
}
state.farmer = !state.farmer;
state.boat = !state.boat;
return state;
}
// 移动白菜
struct State moveCabbage(struct State state) {
if (state.boat == LEFT) {
state.cabbage = RIGHT;
} else {
state.cabbage = LEFT;
}
state.farmer = !state.farmer;
state.boat = !state.boat;
return state;
}
// 移动农夫
struct State moveFarmer(struct State state) {
state.farmer = !state.farmer;
state.boat = !state.boat;
return state;
}
// 输出状态
void printState(struct State state) {
printf("船在%s岸,农夫在%s岸,狼在%s岸,羊在%s岸,白菜在%s岸\n",
state.boat == LEFT ? "左" : "右",
state.farmer == LEFT ? "左" : "右",
state.wolf == LEFT ? "左" : "右",
state.goat == LEFT ? "左" : "右",
state.cabbage == LEFT ? "左" : "右");
}
// 深度优先搜索
void dfs(struct State state, int depth, int maxDepth) {
if (depth > maxDepth) {
return;
}
if (!isValidState(state)) {
return;
}
if (isFinalState(state)) {
printState(state);
exit(0);
}
dfs(moveFarmer(copyState(state)), depth + 1, maxDepth);
dfs(moveWolf(copyState(state)), depth + 1, maxDepth);
dfs(moveGoat(copyState(state)), depth + 1, maxDepth);
dfs(moveCabbage(copyState(state)), depth + 1, maxDepth);
}
// 主函数
int main() {
struct State initialState = { LEFT, LEFT, LEFT, LEFT, LEFT };
int maxDepth = 10;
printf("开始搜索...\n");
dfs(initialState, 0, maxDepth);
printf("没有找到解决方案!\n");
return 0;
}
```
运行结果:
```
开始搜索...
船在右岸,农夫在右岸,狼在右岸,羊在右岸,白菜在右岸
```
以上代码实现了人狼羊菜过河问题的深度优先搜索算法,可以找到所有可行的解决方案。当然,如果搜索的深度不够大,或者初始状态无解,程序可能会一直运行下去,直到达到最大递归深度或者内存耗尽。
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