LeetCode 773. Sliding Puzzle--BFS

最新推荐文章于 2024-11-26 00:29:42 发布

原创最新推荐文章于 2024-11-26 00:29:42 发布 · 609 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#BFS #leetcode #c #算法

LeetCode 专栏收录该内容

29 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索(BFS)解决2x3滑动拼图问题的方法，目标是将拼图从任意初始状态转换为标准解[[1,2,3],[4,5,0]]。通过具体实例演示了如何找到达到目标状态所需的最小移动次数。

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;
    }
};