不相交集圖學習

不相交集是 Kruskal 最小生成樹中使用的資料結構。
這種資料結構允許我們建立兩個或多個節點的並集。
它讓我們確定兩個節點是否屬於 not 圖的同一組成部分。
時間複雜度為 O(4alpha)（如果我們使用的是路徑壓縮，否則它將是對數），這是已被證明的恆定時間複雜度。

更多資訊請參閱

class Main{
    int parent[]  = new int[100000];
    int rank[] = new int[100000];
    void makeSet(){
        for(int i=1;i<=n;i++){
            parent[i] = i; // initially parent of node i is i itself
            rank[i] =0;// intially rank of all the nodes are 0
        }
    }

    int findParent(int node){
        if(node == parent[node]) return node; // ie if the parent of 'node' is 'node' itself then just return the node.
        //return findParent(parent[node]);// else recursively call findParent to get the parent of this union.
        // in order to path conpress (making sure that every node is connected to its parent in the given union)
        return parent[node]  = findParent(parent[node]);
        //example : 7->6->4->3 , here 3 is the parent of this union, so in order to get the parent of 7 which is 3 we can path compress it. like 7->3,6->3,4->3 etc.
    }
    void union(int u, int v){
        u = findParent(u);
        v = findParent(v);
        if(rank[u] < rank[v]) parent[u] =v;
        else if (rank[u] > rank[v]) parent[v] = u;
        else parent[u] =v; // or parent[v] = u;
        // here u and v had the same rank
        //and now v became parent of u hence its rank will increase
        rank[v]++;
    }

    public static void main(String args[]) throws Exception{
        Main m = new Main();
        //if we where given no of nodes as n
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        while(n>0){
            int u = n--;
            int v = n--;
            m.union(u,v);// this will create union of n nodes
        }
        // if 2 and 3 belong to the same component or not find out
        if(findParent(2)! = findParent(3)) System.out.println("Different component");
        else System.out.println("Same component");
    }
}

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