leetcode 694. Number of Distinct Islands

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:
11000
11000
00011
00011
Given the above grid map, return 1.
Example 2:
11011
10000
00001
11011
Given the above grid map, return 3.

Notice that:
11
1
and
 1
11
are considered different island shapes, because we do not consider reflection / rotation.

题目大意，求不同小岛的数量
矩阵中每个点都可以表示成 i*col + j的形式，比如第一个样例的两个岛可以写成(0,1,5,6)(13,14,18,19) 那么如何判断这两个岛是否相等呢
把这两个数组平移到0行y列就可以了。(注意y是小岛最小的左边界)。 就是减去每个数组x的最小值和y的最小值。 那么(0,1,5,6)和(13,14,18,19)-13就相等了，所以他们是一个形状的岛

class Solution {
public:
    int dir[4][2] = {1,0,0,1,0,-1,-1,0};
    int n, m, ans, cntx, cnty;
    map<pair<int,int>, int>mp;
    vector<int> v;
    vector<set<int> > w;
    void dfs(int x, int y, vector<vector<int>>& grid) {
        mp[{x, y}] = 1;
        v.push_back(x * m + y);
        cntx = min(cntx, x);
        cnty = min(cnty, y);
        for (int i = 0; i < 4; ++i) {
            int nx = x + dir[i][0];
            int ny = y + dir[i][1];
            if (nx < n && ny < m && nx >=0 && ny >= 0 && !mp[{nx,ny}] && (grid[nx][ny] == 1)) {
                dfs(nx, ny, grid);
            }
        }
    }
    int numDistinctIslands(vector<vector<int>>& grid) {
        n = grid.size();
        if (n == 0) return 0;
        m = grid[0].size();
        mp.clear();
        ans = 0;
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (mp[{i,j}] == 0 && grid[i][j] == 1) {
                    cntx = cnty = 100000000;
                    v.clear();
                    dfs(i, j, grid);
                    cout << "cnt " << cntx <<" " << cnty << endl;
                    set<int>se;
                    for(auto xp : v) {
                        se.insert(xp-cntx*m-cnty);
                    }
                    w.push_back(se);
                }
            }
        }
        if (w.size()) ans = 1;
        else return 0;
        sort(w.begin(), w.end());
        for (int i = 1; i < w.size(); ++i) {
            if (w[i] != w[i-1]) ans++;
        }
        return ans;
    }
};
/*
011
111
000
111
110
*/
/*
1 4 5 2 3 
9 12 13 10 11 
*/