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 they have the same shape, or have the same shape after rotation (90, 180, or 270 degrees only) or reflection (left/right direction or up/down direction).
Example 1:
11000
10000
00001
00011
Given the above grid map, return 1.
Notice that:
11
1
and
1
11
are considered same island shapes. Because if we make a 180 degrees clockwise rotation on the first island, then two islands will have the same shapes.
Example 2:
11100
10001
01001
01110
Given the above grid map, return 2.
Here are the two distinct islands:
111
1
and
1
1
Notice that:
111
1
and
1
111
are considered same island shapes. Because if we flip the first array in the up/down direction, then they have the same shapes.
Note: The length of each dimension in the given grid does not exceed 50.
这道题的难度比较大，涉及到的细节也很多，是一道很好的题。
注意几个细节：
1. set 由红黑树实现，unordered_set由哈希表实现，所以想要给vector去重，要用set。
2. for(auto l : s) ... 这个时候l是一个value，而for(auto &l : s) ... 这个时候l是一个reference。
3. 上下左右翻转，旋转90，180，270，对于一个(x,y)来讲正好对应8种情况。((+/-)(x/y),(+/-)())，可以自己推一遍。
4. 在norm函数里，有两次排序，这两次排序至关重要，第一次是对每一个翻转情况中的点进行排序，这一次排序的目的是为了让
两个ISLAND对应的翻转中最小的点排到第一个来，这样做的目的是因为只有这样，之后两个ISLAND以该点作为原点进行计算时，
两个ISLAND才能得到一样的点的序列。
第二次排序是为了在所有可能中拿点序最小的一个，拿最大也可以，只要把这八个可能用一个来表示就行了，然后插入集合中，利用
set进行去重。
好题。
#include "header.h"
#include <unordered_set>
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (remove_cv<remove_reference<decltype(b)>::type>::type i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define PRINT1D(x) do {REP(i,(x).size()) cout << x[i] << " "; cout << endl;} while 0;
#define PRINT1D(X) \
{ \
REP(i, (X).size()){ \
cout << (X)[i] << " "; \
} \
cout << endl; \
} \
#define PRINT2D(X) \
{ \
REP(i, (X).size()){ \
REP(j,(X)[0].size()){ \
cout << (X)[i][j] << " "; \
} \
cout << endl; \
} \
cout << endl; \
} \
class Solution {
private:
unordered_map<int, vector<pair<int,int> > > map;
public:
void dfs(vector<vector<int>>& grid, int r, int c, int cnt){
grid[r][c] = 0;
map[cnt].push_back({r,c});
int n = grid.size();
int m = grid[0].size();
if(r < n && r >= 0 && c < m && c >= 0 && grid[r][c] == 1){
dfs(grid, r+1, c, cnt);
dfs(grid, r, c+1, cnt);
dfs(grid, r-1, c, cnt);
dfs(grid, r, c-1, cnt);
}
}
vector<pair<int,int>> norm(vector<pair<int,int>> originalshape){
vector<vector<pair<int,int>>> s(8);
//sort(ALL(originalshape));
for(auto p : originalshape){
int x = p.first, y = p.second;
s[0].push_back({x,y});
s[1].push_back({x,-y});
s[2].push_back({-x,-y});
s[3].push_back({-x,y});
s[4].push_back({y,-x});
s[5].push_back({-y,-x});
s[6].push_back({-x,-y});
s[7].push_back({y,x});
}
//sort(ALL(s));
for(auto &l : s ) sort(ALL(l));
for(auto &l : s){
auto l1st = l[0];
REP(i,l.size()){
l[i].first = l[i].first - l1st.first;
l[i].second = l[i].second - l1st.second;
}
l1st.first = 0;
l1st.second = 0;
}
sort(ALL(s));
return s[0];
};
int numDistinctIslands2(vector<vector<int>>& grid) {
set<vector<pair<int,int>>> s;
int cnt = 0;
REP(i,grid.size()){
REP(j,grid[0].size()){
if(grid[i][j] == 1){
dfs(grid, i, j, ++cnt);
s.insert(norm(map[cnt]));
}
}
}
return s.size();
}
};
int main() {
vector<vector<int>> mtx(4,vector<int>(4,0));
set<vector<int>>s;
Solution s1 = Solution();
s1.numDistinctIslands2(mtx);
}