LeetCode周赛#112 Q3 Most Stones Removed with Same Row or Column（连通分量）

题目来源：https://leetcode.com/contest/weekly-contest-112/problems/most-stones-removed-with-same-row-or-column/

问题描述

 947. Most Stones Removed with Same Row or Column

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

Now, a move consists of removing a stone that shares a column or row with another stone on the grid.

What is the largest possible number of moves we can make?

 

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3

Example 3:

Input: stones = [[0,0]]
Output: 0

Note:

1. 1 <= stones.length <= 1000
2. 0 <= stones[i][j] < 10000

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题意

给定坐标系中若干个点的坐标，如果一个点所在的行或所在列存在别的点，则可以将该点删去。问最多可以删去多少个点。

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思路

如果两点同行或同列，则在两个点之间建立一条边，如此建图，得到图的关联矩阵。问题转化为求图的连通分量，如果求出连通分量，则可以删掉的点为所有连通分量所含的点数-1之和。用bfs求连通分量。

注意：贪心法不断选取度最小的点删去不能得到本题的解。在此给出一个图作为反例：

 

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代码

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int x, y, i, cnt = 0, ans = 0, n = stones.size(), u, v;
        list<int> X[10005], Y[10005];
        list<int> graph[1005];
        bool vis[1005] = {};
        list<int>::iterator iter;
        i = 0;
        for (auto stone : stones)
        {
            x = stone[0];
            y = stone[1];
            X[x].push_back(i);
            Y[y].push_back(i++);
        }
        i = 0;
        for (auto stone : stones)
        {
            x = stone[0];
            y = stone[1];
            iter = graph[i].end();
            iter = graph[i].insert(iter, X[x].begin(), X[x].end());
            graph[i].insert(iter, Y[y].begin(), Y[y].end());
            i++;
        }
        for (i=0; i<n; i++)
        {
            if (!vis[i])
            {
                queue<int> q;
                q.push(i);
                vis[i] = true;
                cnt = 1;
                while (!q.empty())
                {
                    u = q.front();
                    q.pop();
                    for (auto v : graph[u])
                    {
                        if (!vis[v])
                        {
                            q.push(v);
                            vis[v] = true;
                            cnt++;
                        }
                    }
                }
                ans += cnt - 1;
            }
        }
        return ans;
    }
};