技術標籤：LeetCodeC++c++leetcode演算法

Number of Provinces

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3
 

Constraints:

• 1 <= n <= 200
• n == isConnected.length
• n == isConnected[i].length
• isConnected[i][j] is 1 or 0.
• isConnected[i][i] == 1
• isConnected[i][j] == isConnected[j][i]

並查集：

​ 這是一道明顯到不能再明顯的並查集的題目。

​ 有 n 個城市，其中一些彼此相連，另一些沒有相連。如果城市 a 與城市 b 直接相連，且城市 b 與城市 c 直接相連，那麼城市 a 與城市 c 間接相連。

​ 省份是一組直接或間接相連的城市，組內不含其他沒有相連的城市。

​ 使用鄰接矩陣來儲存無向圖。

​ 我的程式碼中用類實現了並查集的合併和查詢操作，查詢操作完成了路徑壓縮。

class DSU
{
    private:
        int n;
        vector<long> f;
    public:
        DSU(int N){
            n = N;
            f.resize(n, 1);
            for(int i=0; i<n; i++)
                f[i] = i;
        }
        int find(int x){
            if(x != f[x])
                f[x] = find(f[x]);
            return f[x];
        }
        void unionset(int x, int y){
            x = find(x);
            y = find(y);
            if(x == y)
                return;
            f[x] = y;
        }
};

class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        DSU dsu(n);
        set<int> s;
        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++)
                if(isConnected[i][j])
                    dsu.unionset(i, j);
        }
        for(int i=0; i<n; i++)
            s.insert(dsu.find(i));
        return s.size();
    }
};