解题思路:
3种扩展方式而已
参考代码:
#include <iostream> #include <string> #include <algorithm> #include <unordered_map> #include <queue> #define x first #define y second using namespace std; unordered_map<string, int> dist; unordered_map<string, pair<char, string>> pre; string st = "12345678", ed, res; string move0(string s) { for (int i = 0; i < 4; i++) swap(s[i], s[7-i]); return s; } string move1(string s) { for (int i = 0; i < 3; i++) swap(s[3], s[i]); for (int i = 4; i < 7; i++) swap(s[i], s[i+1]); return s; } string move2(string s) { swap(s[1], s[2]), swap(s[5], s[6]), swap(s[1], s[5]); return s; } void bfs() { queue<string> q; q.push(st); dist[st] = 0; while (!q.empty()) { string t = q.front(); q.pop(); if (t == ed) return ; string m[3]; m[0] = move0(t); m[1] = move1(t); m[2] = move2(t); for (int i = 0; i < 3; i++) if (!dist.count(m[i])) { q.push(m[i]); dist[m[i]] = dist[t] + 1; pre[m[i]] = {'A' + i, t}; // second经过first操作到达m[i] } } } int main() { for (int i = 1; i <= 8; i++) { char ch; cin >> ch; ed += ch; } bfs(); cout << dist[ed] << endl; if (dist[ed]) { while (ed != st) { res += pre[ed].x; ed = pre[ed].y; } reverse(res.begin(), res.end()); cout << res << endl; } return 0; }
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