【启发式搜索】八数码问题

首先就是f(n)=g(n)+h(n)这个h(n)就是估价函数,然后每次更新一下g(n)然后用康托展开,搞一下判重就好了。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int MAXVIS = 1000000;
struct State{
    int s[9];
    int step;
    int h, g;
    bool operator < (const State& s) const {
        return (g+h)>(s.g+s.h);
    }
    bool operator > (const State& s) const {
        return (g+h)<(s.g+s.h);
    }
}st, ed;
priority_queue<State> que;
int _pow[9]; pair<int, int> show[9];
int fx[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
bool vis[MAXVIS+5];
int hash(int s[]){
    int ret = 0, count =0 ;
    for(int i=0;i<9;i++){
        count = 0;
        for(int j=i+1;j<9;j++)
            count += int(s[j]<s[i]);
        ret += count * _pow[8-i];
    }
    return ret;
}
int _abs(int u){return u>0?u:-u;}
int _h(int s[]){
    int ret = 0;
    for(int i=0;i<9;i++) if(s[i])
        ret += _abs(i/3-show[s[i]].first) + _abs(i%3-show[s[i]].second);
    return ret;
}
bool check(State s){
    for(int i=0;i<9;i++)
        if(s.s[i] != ed.s[i])
            return false;
    return true;
}
bool check_pos(int a, int b){
    if(a > 2 || b > 2 || a < 0 || b < 0) 
        return false;
    return true;
}
void solve(){
    //printf("666");
    State tmp=st; State t2;
    int id=hash(st.s);
    vis[id] = true;
    tmp.h = _h(st.s); tmp.h = 0;
    que.push(tmp);
    while(!que.empty()){
        //printf("6");
        tmp = que.top(); que.pop();
        //printf("6");
        if(check(tmp)) break;
        int zx=0, zy=0, pos=0;
        while(tmp.s[pos]) pos++;
        zx = pos / 3; zy = pos % 3;
        for(int i=0;i<4;i++) {
            int nzx = zx + fx[i][0], nzy = zy + fx[i][1];
            //printf("6");
            if(check_pos(nzx, nzy)){
                t2 = tmp;
                swap(t2.s[pos], t2.s[nzx*3+nzy]);
                id = hash(t2.s);
                if(vis[id]) continue;
                vis[id] = true;
                t2.g++; t2.step++;
                t2.h = _h(t2.s);
                que.push(t2);
            }
        }
    }
    if(check(tmp)) printf("%d\n", tmp.step);
    else printf("-1\n");
}
int main(){
    _pow[0] = 1;
    for(int i=1;i<=8;i++) _pow[i] = _pow[i-1] * i;
    for(int i=0;i<9;i++) scanf("%d", &st.s[i]);
    for(int i=0;i<9;i++){
        scanf("%d", &ed.s[i]);
        show[ed.s[i]] = make_pair(i/3, i%3);
    }
    if(check(st)) printf("0\n");
    else solve();

    return 0;
}

转载于:https://www.cnblogs.com/JeremyGJY/p/5921688.html

### 启发式搜索解决八数码问题的C语言实现 以下是基于启发式搜索算法(如A*算法)解决八数码问题的一个完整C语言实现代码。该代码利用结构体存储节点信息,并通过计算启发函数 \( h(x) \) 来评估当前状态与目标状态的距离。 #### 代码实现 ```c #include <stdio.h> #include <stdlib.h> #define MAX_NODES 100000 typedef struct { int state[3][3]; // 当前状态 int g; // 到达此状态所需的步数 (g(n)) int h; // 启发值,即到目标状态的估计距离 (h(n)) } Node; // 计算曼哈顿距离作为启发值 h(n) int calculateHeuristic(int current_state[3][3], int goal_state[3][3]) { int distance = 0; for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { if (current_state[i][j] != 0 && current_state[i][j] != goal_state[i][j]) { for (int k = 0; k < 3; ++k) { for (int l = 0; l < 3; ++l++) { if (goal_state[k][l] == current_state[i][j]) { distance += abs(k - i) + abs(l - j); } } } } } } return distance; } // 打印当前状态 void printState(int state[3][3]) { for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { printf("%d ", state[i][j]); } printf("\n"); } printf("\n"); } // 复制状态 void copyState(int dest[3][3], int src[3][3]) { for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { dest[i][j] = src[i][j]; } } } // 寻找空白位置 void findBlankPosition(int state[3][3], int *x, int *y) { for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { if (state[i][j] == 0) { *x = i; *y = j; return; } } } } // A* 算法求解八数码问题 int solvePuzzle(Node start_node, Node goal_node) { Node nodes[MAX_NODES]; int visited_states[MAX_NODES] = {0}; int num_nodes = 0; nodes[num_nodes++] = start_node; while (num_nodes > 0) { int min_index = -1; int min_f = INT_MAX; // 查找 f 值最小的节点 for (int i = 0; i < num_nodes; ++i) { int f_value = nodes[i].g + nodes[i].h; if (f_value < min_f) { min_f = f_value; min_index = i; } } if (min_index == -1) break; Node current_node = nodes[min_index]; // 如果找到目标状态,则返回成功 if (current_node.h == 0) { printf("Solution found!\n"); return 1; } // 移除当前节点并标记已访问 for (int i = min_index; i < num_nodes - 1; ++i) { nodes[i] = nodes[i + 1]; } --num_nodes; int blank_x, blank_y; findBlankPosition(current_node.state, &blank_x, &blank_y); // 尝试移动空白格子的方向 int dx[] = {-1, 1, 0, 0}; int dy[] = {0, 0, -1, 1}; for (int dir = 0; dir < 4; ++dir) { int new_blank_x = blank_x + dx[dir]; int new_blank_y = blank_y + dy[dir]; if (new_blank_x >= 0 && new_blank_x < 3 && new_blank_y >= 0 && new_blank_y < 3) { Node child_node; copyState(child_node.state, current_node.state); int temp = child_node.state[new_blank_x][new_blank_y]; child_node.state[new_blank_x][new_blank_y] = 0; child_node.state[blank_x][blank_y] = temp; child_node.g = current_node.g + 1; child_node.h = calculateHeuristic(child_node.state, goal_node.state); // 检查是否已经访问过该状态 int is_visited = 0; for (int i = 0; i < num_nodes; ++i) { if (child_node.h == nodes[i].h && child_node.g == nodes[i].g) { is_visited = 1; break; } } if (!is_visited) { nodes[num_nodes++] = child_node; } } } } return 0; } int main() { Node start_node, goal_node; printf("Enter the initial state of the puzzle:\n"); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { scanf("%d", &start_node.state[i][j]); } } printf("Enter the goal state of the puzzle:\n"); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { scanf("%d", &goal_node.state[i][j]); } } start_node.g = 0; start_node.h = calculateHeuristic(start_node.state, goal_node.state); if (solvePuzzle(start_node, goal_node)) { printf("The puzzle has been solved.\n"); } else { printf("No solution exists.\n"); } return 0; } ``` 上述代码实现了A*算法来解决八数码问题[^1]。它定义了一个 `Node` 结构体用于保存每个状态的信息,包括当前的状态矩阵、到达该状态的成本以及启发式的估值。通过不断扩展具有最低总成本 \( f(n) = g(n) + h(n) \) 的节点,最终可以找到最优路径。 ---
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