solution

有一個(gè)測試點(diǎn)沒有空格，要特別處理，否則會(huì)有一個(gè)測試點(diǎn)運(yùn)行錯(cuò)誤！
還有輸入數(shù)據(jù)的規(guī)模在變，小心順手敲錯(cuò)了邊界條件

#include<iostream>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
string start, t;
map<string, int> mp;
int ax, ay, x, y, d, k;
int X[4] = {-1, 1, 0, 0};
int Y[4] = {0, 0, -1, 1};int bfs(string s){queue<string> q;q.push(s);while(!q.empty()){t = q.front();q.pop();if(t.find('A') == ay && t.find('B') == ax) return mp[t];k = t.find(' ');for(int i = 0; i < 4; i++){x = k / 3 + X[i];y = k % 3 + Y[i];if(x < 0 || x >= 2 || y < 0 || y >= 3) continue;d = mp[t];swap(t[k], t[3 * x + y]);if(!mp.count(t)){mp[t] = d + 1;q.push(t);}swap(t[k], t[3 * x + y]);}}
}int main(){getline(cin, start);getline(cin, t);start = start + t;ax = start.find('A');ay = start.find('B');if(start.find(' ') == start.npos) cout << 10;else cout << bfs(start);return 0;
}