947. Most Stones Removed with Same Row or Column --- M ！！

 On a 2D plane, we place n stones at some integer coordinate points. Each coordinate point may have at most one stone.

A stone can be removed if it shares either the same row or the same column as another stone that has not been removed.

Given an array stones of length n where stones[i] = [xi, yi] represents the location of the ith stone, return the largest possible number of stones that can be removed.

 

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5
Explanation: One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Explanation: One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.

Example 3:

Input: stones = [[0,0]]
Output: 0
Explanation: [0,0] is the only stone on the plane, so you cannot remove it.

 

Constraints:

• 1 <= stones.length <= 1000
• 0 <= xi, yi <= 104
• No two stones are at the same coordinate point.

A:

仔细看题目。

发现其实是计算 number of connected component. 这就简单了，直接染色就行

class Solution {

public:

int removeStones(vector<vector<int>>& stones) {

unordered_map<int, unordered_set<int>> rowM;

unordered_map<int, unordered_set<int>> colM;

unordered_map<int, bool> visited;

for (auto stone : stones) {

int x = stone[0], y = stone[1];

visited[x * MULTIPLIER + y] = false;

rowM[x].insert(y);

colM[y].insert(x);

}

int componentCount = 0;

// converted to # of connected component (same row/colom as edge)

for (auto stone : stones) {

int x = stone[0], y = stone[1];

if (not visited[x * MULTIPLIER + y]) {

dfs(rowM, colM, visited, x, y);

componentCount++;

}

}

return stones.size() - componentCount;

}

private:

int MULTIPLIER = 10001; // using it to do hashing of {row, col}

void dfs(unordered_map<int, unordered_set<int>>& rowM,

unordered_map<int, unordered_set<int>>& colM,

unordered_map<int, bool>& visited, int x, int y) {

visited[x * MULTIPLIER + y] = true;

for (auto col : rowM[x]) {

if (not visited[x * MULTIPLIER + col]) {

dfs(rowM, colM, visited, x, col);

}

}

for (auto row : colM[y]) {

if (not visited[row * MULTIPLIER + y]) {

dfs(rowM, colM, visited, row, y);

}

}

}

};

Learned:

一开始被chatgpt给骗了。  就用了unordered_map<pair<int,int>, bool> 格式。但是编译器没有默认的把 pair<int, int> 给hash的算法。 需要自己定义。 因此就自己做了个乘数 row * MULTIPLIER +y 来做hashing value