LEETCODE 5257. 统计封闭岛屿的数目 Number of Closed Islands

 1 class Solution {
 2 public:
 3     
 4 int MAX_NUM;
 5 vector<vector<int>> field;
 6 vector<int> fa;
 7 vector<int> addx;
 8 vector<int> addy;
 9 void init(int n)
10 {
11     for (int i = 0; i <= n; i++)
12         fa[i] = i;
13 }
14 int get(int x)
15 {
16     return fa[x] == x ? x : fa[x] = get(fa[x]);//路径压缩,防止链式结构
17 }
18 void merge(int x, int y)
19 {
20     fa[get(x)] = get(y);
21 }
22 //================================================
23 void check(int x, int y, vector<vector<int>>& grid)
24 {
25     for (int i = 0; i < 4; i++) {
26         int newx = x + addx[i];
27         int newy = y + addy[i];
28 
29         if (newx >= 0 && newx < grid.size() && newy >= 0 && newy < grid[0].size()
30             && grid[newx][newy] == 0)
31         {
32             int idx = x * grid[0].size() + y;
33             int anotherIdx = newx * grid[0].size() + newy;
34             merge(idx, anotherIdx);
35         }
36     }
37 }
38 
39 int closedIsland(vector<vector<int>>& grid) {
40     MAX_NUM = 110;
41     field = vector<vector<int>>(MAX_NUM, vector<int>(MAX_NUM));
42     fa = vector<int>(MAX_NUM*MAX_NUM + 1);
43     init(MAX_NUM*MAX_NUM);
44     addx = vector<int>{ 1,-1,0,0 };
45     addy = vector<int>{ 0,0,-1,1 };
46     for (int i = 0; i < grid.size(); i++) {
47         for (int j = 0; j < grid[0].size(); j++) {
48             if (grid[i][j] == 0) {
49                 check(i, j, grid);
50             }
51         }
52     }
53 
54     set<int> s;
55 
56     for (int i = 0; i < grid.size(); i++) {
57         for (int j = 0; j < grid[0].size(); j++) {
58             if (grid[i][j] == 0) {
59                 int idx = i * grid[0].size() + j;
60                 s.insert(get(idx));
61             }
62         }
63     }
64 
65     //从统计的并查集 删除与边沿有关的陆地
66     for (int i = 0; i < grid.size(); i++) {
67         for (int j = 0; j < grid[0].size(); j++) {
68             if (grid[i][j] == 0 && (i == 0 || i == grid.size() - 1 || j == 0 || j == grid[0].size() - 1)) {
69                 int idx = i * grid[0].size() + j;
70                 s.erase(get(idx));
71             }
72         }
73     }
74 
75     return s.size();
76 }
77 
78 };