AcWing845.八數碼

題解

還是一道bfs標準題，每次存入4個方向的狀態，直到找到最終狀態

#include <iostream>
#include <cstdio>
#include <unordered_map>
#include <queue>

using namespace std;

int dir[4][2] = {1,0 ,-1,0 ,0,1, 0,-1};
queue<string> q;
unordered_map<string, int> dist;

int BFS(string start)
{
    string end = "12345678x";
    q.push(start);
    dist[start] = 0;
    int x, y, pos, xx, yy, distance;
    while(q.size())
    {
        string t = q.front();
        distance = dist[t];
        q.pop();
        if(t == end) return dist[t];
        pos = t.find("x");
        x = pos / 3, y = pos % 3;
        for(int i = 0; i < 4; ++i)
        {
            xx = x + dir[i][0], yy = y + dir[i][1];
            if( xx >= 0 && xx < 3 && yy >= 0 && yy < 3)
            {
                swap(t[xx * 3 + yy], t[pos]); 
                if(!dist.count(t))
                {
                    q.push(t), dist[t] = distance + 1;
                }
                swap(t[xx * 3 + yy], t[pos]); //恢復原形，用於下次計算
            }
        }
    }
    return -1;
}

int main()
{
    string s;
    char c;
    for(int i = 0; i < 9; ++i)
        cin >> c, s += c;
    cout << BFS(s) << endl;
    return 0;
}