标签：sel out type rar eal 一个 als lin site
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]]
Input: board = [[3,2,4],[1,5,0]] Output: 14
Note:
board will be a 2 x 3 array as described above.board[i][j] will be a permutation of [0, 1, 2, 3, 4, 5].给定2行3列的矩阵board，包含数字0 - 5，求将其恢复为[[1,2,3],[4,5,0]]的状态最少需要移动多少次。
拼图问题，其实就是八数码问题，给定一个可移动的数字，该数字每次只能朝四个方向移动，问最少经过多少次移动能够完成拼图（给定的序列）。刚开始看到的时候完全没思路，后来想到了用BFS来解，BFS相当于一种暴力搜索，每次去搜索所有可能的结果（对本题来说就是三个方向），直到满足条件或退出。事实上看了一些博客了解到更好的解法是A*，这里先不考虑A*，在后面的寻路算法的实现中会写A*的实现。关于BFS的实现，自我感觉自己的写法应该是比较优秀的写法，相对于网上的很多实现来讲，更加简洁，也符合C++的标准。
解法：BFSpublic int slidingPuzzle(int[][] board) {
Set<String> seen = new HashSet<>(); // used to avoid duplicates
String target = "123450";
// convert board to string - initial state.
String s = Arrays.deepToString(board).replaceAll("\\[|\\]|,|\\s", "");
Queue<String> q = new LinkedList<>(Arrays.asList(s));
seen.add(s); // add initial state to set.
int ans = 0; // record the # of rounds of Breadth Search
while (!q.isEmpty()) { // Not traverse all states yet?
// loop used to control search breadth.
for (int sz = q.size(); sz > 0; --sz) {
String str = q.poll();
if (str.equals(target)) { return ans; } // found target.
int i = str.indexOf(‘0‘); // locate ‘0‘
int[] d = { 1, -1, 3, -3 }; // potential swap displacements.
for (int k = 0; k < 4; ++k) { // traverse all options.
int j = i + d[k]; // potential swap index.
// conditional used to avoid invalid swaps.
if (j < 0 || j > 5 || i == 2 && j == 3 || i == 3 && j == 2) { continue; }
char[] ch = str.toCharArray();
// swap ch[i] and ch[j].
char tmp = ch[i];
ch[i] = ch[j];
ch[j] = tmp;
s = String.valueOf(ch); // a new candidate state.
if (seen.add(s)) { q.offer(s); } //Avoid duplicate.
}
}
++ans; // finished a round of Breadth Search, plus 1.
}
return -1;
}
Java:
public int slidingPuzzle(int[][] board) {
String target = "123450";
String start = "";
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
start += board[i][j];
}
}
HashSet<String> visited = new HashSet<>();
// all the positions 0 can be swapped to
int[][] dirs = new int[][] { { 1, 3 }, { 0, 2, 4 },
{ 1, 5 }, { 0, 4 }, { 1, 3, 5 }, { 2, 4 } };
Queue<String> queue = new LinkedList<>();
queue.offer(start);
visited.add(start);
int res = 0;
while (!queue.isEmpty()) {
// level count, has to use size control here, otherwise not needed
int size = queue.size();
for (int i = 0; i < size; i++) {
String cur = queue.poll();
if (cur.equals(target)) {
return res;
}
int zero = cur.indexOf(‘0‘);
// swap if possible
for (int dir : dirs[zero]) {
String next = swap(cur, zero, dir);
if (visited.contains(next)) {
continue;
}
visited.add(next);
queue.offer(next);
}
}
res++;
}
return -1;
}
private String swap(String str, int i, int j) {
StringBuilder sb = new StringBuilder(str);
sb.setCharAt(i, str.charAt(j));
sb.setCharAt(j, str.charAt(i));
return sb.toString();
}
def slidingPuzzle(self, board):
self.goal = [[1,2,3], [4,5,0]]
self.score = [0] * 6
self.score[0] = [[3, 2, 1], [2, 1, 0]]
self.score[1] = [[0, 1, 2], [1, 2, 3]]
self.score[2] = [[1, 0, 1], [2, 1, 2]]
self.score[3] = [[2, 1, 0], [3, 2, 1]]
self.score[4] = [[1, 2, 3], [0, 1, 2]]
self.score[5] = [[2, 1, 2], [1, 0, 1]]
heap = [(0, 0, board)]
closed = []
while len(heap) > 0:
node = heapq.heappop(heap)
if node[2] == self.goal:
return node[1]
elif node[2] in closed:
continue
else:
for next in self.get_neighbors(node[2]):
if next in closed: continue
heapq.heappush(heap, (node[1] + 1 + self.get_score(next), node[1] + 1, next))
closed.append(node[2])
return -1
def get_neighbors(self, board):
res = []
if 0 in board[0]:
r, c = 0, board[0].index(0)
else:
r, c = 1, board[1].index(0)
for offr, offc in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
if 0 <= r + offr < 2 and 0 <= c + offc < 3:
board1 = copy.deepcopy(board)
board1[r][c], board1[r+offr][c+offc] = board1[r+offr][c+offc], board1[r][c]
res.append(board1)
return res
def get_score(self, board):
score = 0
for i in range(2):
for j in range(3):
score += self.score[board[i][j]][i][j]
return score
class Solution(object):
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
step = 0
board = tuple(map(tuple, board))
q = [board]
memo = set([board])
while q:
q0 = []
for b in q:
if b == ((1,2,3), (4,5,0)): return step
for x in range(2):
for y in range(3):
if b[x][y]: continue
for dx, dy in zip((1, 0, -1, 0), (0, 1, 0, -1)):
nx, ny = x + dx, y + dy
if 0 <= nx < 2 and 0 <= ny < 3:
nb = list(map(list, b))
nb[nx][ny], nb[x][y] = nb[x][y], nb[nx][ny]
nb = tuple(map(tuple, nb))
if nb not in memo:
memo.add(nb)
q0.append(nb)
q = q0
step += 1
return -1
Python:
# Time: O((m * n) * (m * n)!)
# Space: O((m * n) * (m * n)!)
import heapq
import itertools
# A* Search Algorithm
class Solution(object):
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
def dot(p1, p2):
return p1[0]*p2[0]+p1[1]*p2[1]
def heuristic_estimate(board, R, C, expected):
result = 0
for i in xrange(R):
for j in xrange(C):
val = board[C*i + j]
if val == 0: continue
r, c = expected[val]
result += abs(r-i) + abs(c-j)
return result
R, C = len(board), len(board[0])
begin = tuple(itertools.chain(*board))
end = tuple(range(1, R*C) + [0])
expected = {(C*i+j+1) % (R*C) : (i, j)
for i in xrange(R) for j in xrange(C)}
min_steps = heuristic_estimate(begin, R, C, expected)
closer, detour = [(begin.index(0), begin)], []
lookup = set()
while True:
if not closer:
if not detour:
return -1
min_steps += 2
closer, detour = detour, closer
zero, board = closer.pop()
if board == end:
return min_steps
if board not in lookup:
lookup.add(board)
r, c = divmod(zero, C)
for direction in ((-1, 0), (1, 0), (0, -1), (0, 1)):
i, j = r+direction[0], c+direction[1]
if 0 <= i < R and 0 <= j < C:
new_zero = i*C+j
tmp = list(board)
tmp[zero], tmp[new_zero] = tmp[new_zero], tmp[zero]
new_board = tuple(tmp)
r2, c2 = expected[board[new_zero]]
r1, c1 = divmod(zero, C)
r0, c0 = divmod(new_zero, C)
is_closer = dot((r1-r0, c1-c0), (r2-r0, c2-c0)) > 0
(closer if is_closer else detour).append((new_zero, new_board))
return min_steps
Python:
# Time: O((m * n) * (m * n)! * log((m * n)!))
# Space: O((m * n) * (m * n)!)
# A* Search Algorithm
class Solution2(object):
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
def heuristic_estimate(board, R, C, expected):
result = 0
for i in xrange(R):
for j in xrange(C):
val = board[C*i + j]
if val == 0: continue
r, c = expected[val]
result += abs(r-i) + abs(c-j)
return result
R, C = len(board), len(board[0])
begin = tuple(itertools.chain(*board))
end = tuple(range(1, R*C) + [0])
end_wrong = tuple(range(1, R*C-2) + [R*C-1, R*C-2, 0])
expected = {(C*i+j+1) % (R*C) : (i, j)
for i in xrange(R) for j in xrange(C)}
min_heap = [(0, 0, begin.index(0), begin)]
lookup = {begin: 0}
while min_heap:
f, g, zero, board = heapq.heappop(min_heap)
if board == end: return g
if board == end_wrong: return -1
if f > lookup[board]: continue
r, c = divmod(zero, C)
for direction in ((-1, 0), (1, 0), (0, -1), (0, 1)):
i, j = r+direction[0], c+direction[1]
if 0 <= i < R and 0 <= j < C:
new_zero = C*i+j
tmp = list(board)
tmp[zero], tmp[new_zero] = tmp[new_zero], tmp[zero]
new_board = tuple(tmp)
f = g+1+heuristic_estimate(new_board, R, C, expected)
if f < lookup.get(new_board, float("inf")):
lookup[new_board] = f
heapq.heappush(min_heap, (f, g+1, new_zero, new_board))
return -1
C++:
class Solution {
public:
int slidingPuzzle(vector<vector<int>>& board) {
int res = 0, m = board.size(), n = board[0].size();
string target = "123450", start = "";
vector<vector<int>> dirs{{1,3}, {0,2,4}, {1,5}, {0,4}, {1,3,5}, {2,4}};
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
start += to_string(board[i][j]);
}
}
unordered_set<string> visited{start};
queue<string> q{{start}};
while (!q.empty()) {
for (int i = q.size() - 1; i >= 0; --i) {
string cur = q.front(); q.pop();
if (cur == target) return res;
int zero_idx = cur.find("0");
for (int dir : dirs[zero_idx]) {
string cand = cur;
swap(cand[dir], cand[zero_idx]);
if (visited.count(cand)) continue;
visited.insert(cand);
q.push(cand);
}
}
++res;
}
return -1;
}
};
C++:
class Solution {
public:
int slidingPuzzle(vector<vector<int>>& board) {
int res = 0;
set<vector<vector<int>>> visited;
queue<pair<vector<vector<int>>, vector<int>>> q;
vector<vector<int>> correct{{1, 2, 3}, {4, 5, 0}};
vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 3; ++j) {
if (board[i][j] == 0) q.push({board, {i, j}});
}
}
while (!q.empty()) {
for (int i = q.size() - 1; i >= 0; --i) {
auto t = q.front().first;
auto zero = q.front().second; q.pop();
if (t == correct) return res;
visited.insert(t);
for (auto dir : dirs) {
int x = zero[0] + dir[0], y = zero[1] + dir[1];
if (x < 0 || x >= 2 || y < 0 || y >= 3) continue;
vector<vector<int>> cand = t;
swap(cand[zero[0]][zero[1]], cand[x][y]);
if (visited.count(cand)) continue;
q.push({cand, {x, y}});
}
}
++res;
}
return -1;
}
};
[LeetCode] 773. Sliding Puzzle 滑动拼图
标签：sel out type rar eal 一个 als lin site
原文地址：https://www.cnblogs.com/lightwindy/p/9834773.html
