//  find 的原始实现
int find(int parent[], int i)
{if (parent[i] == -1)return i;return find(parent, parent[i]);
}// union()的原始实现
void Union(int parent[], int x, int y)
{int xset = find(parent, x);int yset = find(parent, y);parent[xset] = yset;
}

假设有4个元素 0, 1, 2, 3
初始化
0 1 2 3 Do Union(0, 1)1   2   3  /0Do Union(1, 2)2   3   /1/
0Do Union(2, 3)3    /2/1/
0

继续用上面的例子
0 1 2 3 Do Union(0, 1)1   2   3  /0Do Union(1, 2)1    3/  \
0    2Do Union(2, 3)1    /  |  \
0   2   3

假设集合{0, 1, .. 9} 的树表示如下所示：  当调用 find(3)时9/    |    \  4     5      6/     \        /  \0        3     7    8/  \1    2  我们向上查找，找到是这个集合的根节点. 路径压缩就是：直接把 3的 父节点 设置为 9
当下次再查找 1, 2 或 3 时，查找的路径就会变短。9/    /  \    \4    5    6     3 /           /  \   /  \0           7    8  1   2

// 用并查集判断是否存在环
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//图中的边
struct Edge
{int dest,src;
};
//图结构体
struct Graph
{
// V-> 顶点个数, E-> 边的个数int V,E;// 每个图就是 边的集合struct Edge *edge;
};struct subset
{int parent;int rank;
};
//创建一个图
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;
}//使用路径压缩
int find(struct subset subsets[],int i)
{
// 找到 root并使  root 作为 i 的父节点if(subsets[i].parent==i)return i;elsereturn (subsets[i].parent=find(subsets,subsets[i].parent));
}//使用按秩合并
void Union(struct subset subsets[], int x, int y)
{int xroot=find(subsets,x);int yroot=find(subsets,y);// 将深度较小的集合 合并到深度大的集合下面if(subsets[xroot].rank<subsets[yroot].rank)subsets[xroot].parent=yroot;else if(subsets[xroot].rank>subsets[yroot].rank)subsets[yroot].parent=xroot;else//深度一样，任选一个，并增加另一个
    {subsets[yroot].parent=xroot;subsets[xroot].rank++;}
}int isCycle(struct Graph* graph)
{int V=graph->V;int E=graph->E;struct subset *subsets=(struct subset *)malloc(V*sizeof(struct subset));for(int i=0;i<V;i++){subsets[i].parent=i;subsets[i].rank=0;}for(int e=0;e<E;e++){int x=find(subsets,graph->edge[e].dest);int y=find(subsets,graph->edge[e].src);if(x==y)return 1;Union(subsets,x,y);}return 0;
}int main()
{int V=3,E=2;struct Graph *graph=createGraph(V,E);/* 构造以下的图0|  \|    \1-----2 */graph->edge[0].src=0;graph->edge[0].dest=1;graph->edge[1].src=1;graph->edge[1].dest=2;// graph->edge[2].src=0;//graph->edge[2].dest=3;if(isCycle(graph))printf("Graph contains cycle\n");elseprintf("Graph doesn't contains cycle\n");return 0;
}