Union-find and Disjoint Set Union

Disjoint Set Union (DSU) is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. A union-find algorithm is an algorithm that performs two useful operations on such a data structure:

Find: Determine which subset a particular element is in. This can be used for determining if two elements are in the same subset.

Union: Join two subsets into a single subset.

DSU很适合用来定义连通分量之类的数据结构。在此基础上的union-find算法主要就是两件事，1.判断任意一个点属于哪个集合或者分量，2.将两个集合合并。
所以用来在无向图里面找环的时候特别好用。首先建立好图的点和线之间的关系，然后初始化一个全是独立的点与点之间的parent关系。将每个点的母结点初始化为-1，就是说每一个点都是自己的leading pixel。在之后用每一条边来建立parent关系就是将一个连通分量的各个点之间，通过这种parent的寻找联系起来。
当我们将每一条边对应的关系添加进DSU之前，如果这个边的两个点已经在一个集合里了，这个时候就证明有环路形成了。
c语言实现代码如下：

// A union-find algorithm to detect cycle in a graph
#include 
#include 
#include 

// a structure to represent an edge in graph
struct Edge
{
    int src, dest;
};

// a structure to represent a graph
struct Graph
{
    // V-> Number of vertices, E-> Number of edges
    int V, E;

    // graph is represented as an array of edges
    struct Edge* edge;
};

// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E)
{
    struct Graph* graph = 
           (struct Graph*) malloc( sizeof(struct Graph) );
    graph->V = V;
    graph->E = E;

    graph->edge = 
        (struct Edge*) malloc( graph->E * sizeof( struct Edge ) );

    return graph;
}

// A utility function to find the subset of an element i
int find(int parent[], int i)
{
    if (parent[i] == -1)
        return i;
    return find(parent, parent[i]);
}

// A utility function to do union of two subsets 
void Union(int parent[], int x, int y)
{
    int xset = find(parent, x);
    int yset = find(parent, y);
    parent[xset] = yset;
}

// The main function to check whether a given graph contains 
// cycle or not
int isCycle( struct Graph* graph )
{
    // Allocate memory for creating V subsets
    int *parent = (int*) malloc( graph->V * sizeof(int) );

    // Initialize all subsets as single element sets
    memset(parent, -1, sizeof(int) * graph->V);

    // Iterate through all edges of graph, find subset of both
    // vertices of every edge, if both subsets are same, then 
    // there is cycle in graph.
    for(int i = 0; i < graph->E; ++i)
    {
        int x = find(parent, graph->edge[i].src);
        int y = find(parent, graph->edge[i].dest);

        if (x == y)
            return 1;

        Union(parent, x, y);
    }
    return 0;
}

// Driver program to test above functions
int main()
{
    /* Let us create following graph
         0
        |  \
        |    \
        1-----2 */    
    int V = 3, E = 3;
    struct Graph* graph = createGraph(V, E);

    // add edge 0-1
    graph->edge[0].src = 0;
    graph->edge[0].dest = 1;

    // add edge 1-2
    graph->edge[1].src = 1;
    graph->edge[1].dest = 2;

    // add edge 0-2
    graph->edge[2].src = 0;
    graph->edge[2].dest = 2;

    if (isCycle(graph))
        printf( "graph contains cycle" );
    else
        printf( "graph doesn't contain cycle" );

    return 0;
}