LeetCode 773. Sliding Puzzle--BFS

求最短的step这类问题可以用BFS解决，我的博客也有一道类似的题目跳蚱蜢

 

773. Sliding Puzzle

On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square represented by 0.

A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.

The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].

Given a puzzle board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.

Examples:

Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.

Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.

Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation: 5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]

代码：

class Solution {
public:
    int slidingPuzzle(vector<vector<int>>& board) {
        string start = "";
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 3; j++)
                start += board[i][j] + 48;
        queue<pair<string, int> > q;
        q.push(make_pair(start, 0));
        set<string> s;
        s.insert(start);
        string e = "123450";
        while(!q.empty()) {
            string a = q.front().first;
            int step = q.front().second;
            if (a == e) return step;
            q.pop();
            int pos = 0;
            while(a[pos]!='0') pos++;
            int posi[3];
            bool threePos = false;
            if (pos / 3 == 0) posi[0] = pos+3;
            else posi[0] = pos-3;
            if (pos == 0 || pos == 3) posi[1] = pos+1;
            if (pos == 2 || pos == 5) posi[1] = pos-1;
            if (pos == 1 || pos == 4) {
                threePos = true;
                posi[1] = pos+1;
                posi[2] = pos-1;
            }
            for (int i = 0; i < 3; i++) {
                if (i == 2 && !threePos) break;
                string b = a;
                b[pos] = a[posi[i]];
                b[posi[i]] = a[pos];
                if (s.count(b) == 0) {
                    s.insert(b);
                    q.push(make_pair(b, step+1));
                }
            }   
        }
        return -1;
    }
};