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LeetCode 1263. 推箱子（BFS+DFS / 自定义哈希set）
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文章目錄
1. 題目
「推箱子」是一款風靡全球的益智小游戲，玩家需要將箱子推到倉庫中的目標位置。
游戲地圖用大小為 n * m 的網格 grid 表示，其中每個元素可以是墻、地板或者是箱子。
現在你將作為玩家參與游戲，按規則將箱子 'B' 移動到目標位置 'T' ：
- 玩家用字符 'S' 表示，只要他在地板上，就可以在網格中向上、下、左、右四個方向移動。
- 地板用字符 '.' 表示，意味著可以自由行走。
- 墻用字符 '#' 表示，意味著障礙物，不能通行。
- 箱子僅有一個，用字符 'B' 表示。相應地，網格上有一個目標位置 'T'。
玩家需要站在箱子旁邊，然后沿著箱子的方向進行移動，此時箱子會被移動到相鄰的地板單元格。記作一次「推動」。
玩家無法越過箱子。
返回將箱子推到目標位置的最小 推動 次數，如果無法做到，請返回 -1。
示例 1：
輸入：grid
= [["#","#","#","#","#","#"],["#","T","#","#","#","#"],["#",".",".","B",".","#"],["#",".","#","#",".","#"],["#",".",".",".","S","#"],["#","#","#","#","#","#"]]
輸出：
3
解釋：我們只需要返回推箱子的次數。示例
2：
輸入：grid
= [["#","#","#","#","#","#"],["#","T","#","#","#","#"],["#",".",".","B",".","#"],["#","#","#","#",".","#"],["#",".",".",".","S","#"],["#","#","#","#","#","#"]]
輸出：
-1示例
3：
輸入：grid
= [["#","#","#","#","#","#"],["#","T",".",".","#","#"],["#",".","#","B",".","#"],["#",".",".",".",".","#"],["#",".",".",".","S","#"],["#","#","#","#","#","#"]]
輸出：
5
解釋：向下、向左、向左、向上再向上。示例
4：
輸入：grid
= [["#","#","#","#","#","#","#"],["#","S","#",".","B","T","#"],["#","#","#","#","#","#","#"]]
輸出：
-1提示：
1 <= grid
.length
<= 20
1 <= grid
[i
].length
<= 20
grid 僅包含字符
'.', '#', 'S' , 'T', 以及
'B'。
grid 中
'S', 'B' 和
'T' 各只能出現一個。
來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/minimum-moves-to-move-a-box-to-their-target-location
著作權歸領扣網絡所有。商業轉載請聯系官方授權，非商業轉載請注明出處。
2. 解題
2.1 超時解
15 / 18 個通過測試用例
- 優先隊列里存入 box 和 人的4個坐標位置，以及推的次數，推的次數小的優先
- 隊列里存了太多的狀態
class node
{
public:int push
, bi
, bj
, pi
, pj
;node(int n
, int a
, int b
, int c
, int d
){push
= n
;bi
= a
;bj
= b
;pi
= c
;pj
= d
;}
};
struct hashf
{unsigned long long operator()(node v
) const{return 1ULL*v
.push
*pow(21,4)+v
.bi
*pow(21,3)+v
.bj
*pow(21,2)+v
.pi
*21+v
.pj
;}
};
struct eqf
{bool operator()(node v1
, node v2
) const{return v1
.push
==v2
.push
&& v1
.bi
==v2
.bi
&& v1
.bj
==v2
.bj
&& v1
.pi
==v2
.pi
&& v1
.pj
==v2
.pj
;}
};
struct cmp
{bool operator()(node
& a
, node
& b
) const{return a
.push
> b
.push
;}
};
class Solution {
public:int minPushBox(vector
<vector
<char>>& grid
) {int m
= grid
.size(), n
= grid
[0].size(), k
= 21;int targetx
, targety
, boxi
, boxj
, peoplei
, peoplej
, push
= 0, size
;for(int i
= 0; i
< m
; i
++){for(int j
= 0; j
< n
; j
++){if(grid
[i
][j
] == 'S')peoplei
= i
, peoplej
= j
;else if(grid
[i
][j
] == 'B')boxi
= i
, boxj
= j
;else if(grid
[i
][j
] == 'T')targetx
= i
, targety
= j
;}}vector
<vector
<int>> dir
= {{1,0},{0,1},{0,-1},{-1,0}};priority_queue
<node
, vector
<node
>, cmp
> q
; node
state(0, boxi
, boxj
, peoplei
, peoplej
);q
.push(state
);unordered_set
<node
, hashf
, eqf
> vis
;vis
.insert(state
);while(!q
.empty()){int pj
= q
.top().pj
;int pi
= q
.top().pi
;int bj
= q
.top().bj
;int bi
= q
.top().bi
;push
= q
.top().push
;if(bi
==targetx
&& bj
==targety
)return push
;q
.pop();int nextbi
, nextbj
, nextpi
, nextpj
;for(int d
= 0; d
< 4; d
++){nextpi
= pi
+dir
[d
][0];nextpj
= pj
+dir
[d
][1];state
.push
= push
;if(nextpi
==bi
&& nextpj
==bj
){ nextbi
= bi
+dir
[d
][0];nextbj
= bj
+dir
[d
][1];state
.push
++;}else{nextbi
= bi
;nextbj
= bj
;}state
.bi
= nextbi
;state
.bj
= nextbj
;state
.pi
= nextpi
;state
.pj
= nextpj
;if(nextpi
>=0 && nextpi
<m
&& nextpj
>=0 && nextpj
<n
&&nextbi
>=0 && nextbi
<m
&& nextbj
>=0 && nextbj
<n
&&grid
[nextpi
][nextpj
] != '#' && grid
[nextbi
][nextbj
] != '#' &&vis
.find(state
) == vis
.end()){vis
.insert(state
);q
.push(state
);}}}return -1;}
};
2.2 BFS + DFS
- 隊列里面只存能推動箱子的狀態
- 中間到達推箱子的過程，不必記錄到隊列內，采用DFS判斷人的位置能否到達推動箱子的位置
- 不采用優先隊列也可以
class node
{
public:int push
, bi
, bj
, pi
, pj
;node(int n
, int a
, int b
, int c
, int d
){push
= n
;bi
= a
;bj
= b
;pi
= c
;pj
= d
;}
};
struct hashf
{unsigned long long operator()(node v
) const{return 1ULL*v
.bi
*pow(21,3)+v
.bj
*pow(21,2)+v
.pi
*21+v
.pj
;}
};
struct eqf
{bool operator()(node v1
, node v2
) const{return v1
.bi
==v2
.bi
&& v1
.bj
==v2
.bj
&& v1
.pi
==v2
.pi
&& v1
.pj
==v2
.pj
;}
};
struct cmp
{bool operator()(node
& a
, node
& b
) const{return a
.push
> b
.push
;}
};
class Solution {vector
<vector
<int>> dir
= {{1,0},{0,1},{0,-1},{-1,0}};
public:int minPushBox(vector
<vector
<char>>& grid
) {int m
= grid
.size(), n
= grid
[0].size();int targetx
, targety
, boxi
, boxj
, peoplei
, peoplej
, push
= 0, size
;for(int i
= 0; i
< m
; i
++){for(int j
= 0; j
< n
; j
++){if(grid
[i
][j
] == 'S')peoplei
= i
, peoplej
= j
;else if(grid
[i
][j
] == 'B')boxi
= i
, boxj
= j
;else if(grid
[i
][j
] == 'T')targetx
= i
, targety
= j
;}}priority_queue
<node
, vector
<node
>, cmp
> q
; node
state(0, boxi
, boxj
, peoplei
, peoplej
);q
.push(state
);unordered_set
<node
, hashf
, eqf
> vis
;vis
.insert(state
);while(!q
.empty()){int pj
= q
.top().pj
;int pi
= q
.top().pi
;int bj
= q
.top().bj
;int bi
= q
.top().bi
;push
= q
.top().push
;if(bi
==targetx
&& bj
==targety
)return push
;q
.pop();int nextbi
, nextbj
, nextpi
, nextpj
, pushPosx
, pushPosy
;for(int d
= 0; d
< 4; d
++){nextbi
= bi
+dir
[d
][0];nextbj
= bj
+dir
[d
][1];if(nextbi
<0 || nextbi
>=m
|| nextbj
<0 || nextbj
>=n
|| grid
[nextbi
][nextbj
] == '#')continue; nextpi
= bi
;nextpj
= bj
;pushPosx
= bi
-dir
[d
][0];pushPosy
= bj
-dir
[d
][1];if(pushPosx
<0 || pushPosx
>=m
|| pushPosy
<0 || pushPosy
>=n
|| grid
[pushPosx
][pushPosy
] == '#')continue; vector
<vector
<bool>> record(m
, vector
<bool>(n
, false));if(!canReachPushPos(grid
,pi
,pj
,bi
,bj
,pushPosx
,pushPosy
,record
))continue;state
.push
= push
+1;state
.bi
= nextbi
;state
.bj
= nextbj
;state
.pi
= nextpi
;state
.pj
= nextpj
;if(vis
.find(state
) == vis
.end()){vis
.insert(state
);q
.push(state
);}}}return -1;}bool canReachPushPos(vector
<vector
<char>>& grid
, int x0
, int y0
, int bi
, int bj
, int tx
, int ty
, vector
<vector
<bool>>& vis
){vis
[x0
][y0
] = true;if(x0
==tx
&& y0
==ty
) return true;int m
= grid
.size(), n
= grid
[0].size();for(int k
= 0; k
< 4; k
++){int nx
= x0
+dir
[k
][0];int ny
= y0
+dir
[k
][1];if(nx
>=0 && nx
<m
&& ny
>=0 && ny
<n
&& grid
[nx
][ny
] != '#' && !(nx
==bi
&& ny
==bj
) && !vis
[nx
][ny
]){ if(canReachPushPos(grid
,nx
, ny
,bi
,bj
,tx
,ty
, vis
))return true;}}return false;}
};
240 ms 20.4 MB C++
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