1192. Critical Connections in a Network (H)

Critical Connections in a Network (H)

题目

There are n servers numbered from 0 to n-1 connected by undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. Any server can reach any other server directly or indirectly through the network.

A critical connection is a connection that, if removed, will make some server unable to reach some other server.

Return all critical connections in the network in any order.

Example 1:

Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.

Constraints:

• 1 <= n <= 10^5
• n-1 <= connections.length <= 10^5
• connections[i][0] != connections[i][1]
• There are no repeated connections.

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题意

在无向图中找到所有的边，使得删去该边后，存在两结点无法互相到达。

思路

tarjan算法，求所有的"桥"。

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代码实现

Java

class Solution {
    private Map<Integer, List<Integer>> graph;
    private boolean[] visited;
    private int[] dfn;
    private int[] low;
    private int num;

    public List<List<Integer>> criticalConnections(int n, List<List<Integer>> connections) {
        graph = new HashMap<>();
        visited = new boolean[n];
        dfn = new int[n];
        low = new int[n];
        num = 0;

        List<List<Integer>> ans = new ArrayList<>();

        for (List<Integer> edge : connections) {
            int a = edge.get(0);
            int b = edge.get(1);
            graph.putIfAbsent(a, new ArrayList<>());
            graph.putIfAbsent(b, new ArrayList<>());
            graph.get(a).add(b);
            graph.get(b).add(a);
        }

        tarjan(0, -1, ans);

        return ans;
    }

    private void tarjan(int cur, int parent, List<List<Integer>> ans) {
        visited[cur] = true;
        dfn[cur] = num++;
        low[cur] = dfn[cur];

        for (int next : graph.get(cur)) {
            if (!visited[next]) {
                tarjan(next, cur, ans);
                low[cur] = Math.min(low[cur], low[next]);
	              // 满足条件时说明删去改变子结点和父结点将不在同一个连通分量中
                if (low[next] > dfn[cur]) ans.add(Arrays.asList(cur, next));
            } else if (next != parent) {			// 必须排除与父结点的边
                low[cur] = Math.min(low[cur], dfn[next]);
            }
        }
    }
}