[宽度优先搜索] leetcode 773 Sliding Puzzle

problem:https://leetcode.com/problems/sliding-puzzle/

       一看到求最短次数就是宽搜了。用字符串来记录所有状态量即可，找到目标状态量后就退出，我们总能保证找到时用到的步数是最少的。

class Solution {
public:
    int slidingPuzzle(vectorint>>& board) {
        queue<string> q;
        string begin;
        int m = board.size();
        int n = board[0].size();
        for (int i = 0; i < board.size(); i++)
        {
            for (int j = 0; j < board[i].size(); j++)
            {
                begin.push_back(board[i][j] + '0');
            }
        }

        int res = 0;

        q.push(begin);
        bool bFind = false;
        vector<int> dx{ 0,1,0,-1 };
        vector<int> dy{ 1,0,-1,0 };
        unordered_set<string> visit;
        while (!q.empty())
        {
            int size = q.size();
            while (size--)
            {
                string cur = q.front(); q.pop();
                if (cur == "123450")
                {
                    bFind = true;
                    break;
                }
                size_t pos = cur.find('0');
                int i = pos / n;
                int j = pos % n;
                for (int k = 0; k < 4; k++)
                {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if (x >= 0 && y >= 0 && x < m && y < n)
                    {
                        int newPos = x * n + y;
                        string next = cur;
                        swap(next[pos], next[newPos]);
                        if (visit.find(next) == visit.end())
                        {
                            visit.insert(next);
                            q.push(next);
                        }
                    }
                }
            }
            if (bFind) break;
            res++;
        }
        if (!bFind) return -1;
        return res;
    }
};

 

转载于:https://www.cnblogs.com/fish1996/p/11309552.html