Leetcode 1761. Minimum Degree of a Connected Trio in a Graph (图好题)

1761. Minimum Degree of a Connected Trio in a Graph
      Hard

You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edgesi = ui, vi indicates that there is an undirected edge between ui and vi.

A connected trio is a set of three nodes where there is an edge between every pair of them.

The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not.

Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.

Example 1:

Input: n = 6, edges = \[1,2,1,3,3,2,4,1,5,2,3,6]

Output: 3

Explanation: There is exactly one trio, which is 1,2,3. The edges that form its degree are bolded in the figure above.

Example 2:

Input: n = 7, edges = \[1,3,4,1,4,3,2,5,5,6,6,7,7,5,2,6]

Output: 0

Explanation: There are exactly three trios:

1. 1,4,3 with degree 0.
2. 2,5,6 with degree 2.
3. 5,6,7 with degree 2.

Constraints:

2 <= n <= 400

edgesi.length == 2

1 <= edges.length <= n * (n-1) / 2

1 <= ui, vi <= n

ui != vi

There are no repeated edges.

解法1：临接矩阵

class Solution {
public:
    int minTrioDegree(int n, vector<vector<int>>& edges) {
        vector<vector<int>> matrix(n + 1, vector<int>(n + 1));
        vector<int> counter(n + 1);
        int res = INT_MAX;
        for (auto &edge : edges) {
            matrix[min(edge[0], edge[1])][max(edge[0], edge[1])] = 1;
            ++counter[edge[0]];
            ++counter[edge[1]];
        }
        for (auto i = 1; i <= n; i++) {
            for (auto j = i + 1; j <= n; j++) {
                if (matrix[i][j]) {
                    for (auto k = j + 1; k <= n; k++) {
                        if (matrix[i][k] && matrix[j][k]) {
                            res = min(res, counter[i] + counter[j] + counter[k] - 6);
                        }
                    }
                }
            }
        }
        return res == INT_MAX ? -1 : res;
    }
};