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 <stdio.h>#include <stdlib.h>#include <string.h>// a structure to represent an edge in graphstruct Edge{    int src, dest;};// a structure to represent a graphstruct 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 edgesstruct 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 iint 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 notint 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 functionsint 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;}