6-19 Count Connected Components（20 point(s)）

6-19 Count Connected Components（20 point(s)）

Write a function to count the number of connected components in a given graph.

Format of functions:

int CountConnectedComponents( LGraph Graph );

where LGraph is defined as the following:

typedef struct AdjVNode *PtrToAdjVNode; struct AdjVNode{    Vertex AdjV;    PtrToAdjVNode Next;};typedef struct Vnode{    PtrToAdjVNode FirstEdge;} AdjList[MaxVertexNum];typedef struct GNode *PtrToGNode;struct GNode{      int Nv;    int Ne;    AdjList G;};typedef PtrToGNode LGraph;

The function CountConnectedComponents is supposed to return the number of connected components in the undirectedGraph.

Sample program of judge:

#include <stdio.h>#include <stdlib.h>typedef enum {false, true} bool;#define MaxVertexNum 10  /* maximum number of vertices */typedef int Vertex;      /* vertices are numbered from 0 to MaxVertexNum-1 */typedef struct AdjVNode *PtrToAdjVNode; struct AdjVNode{    Vertex AdjV;    PtrToAdjVNode Next;};typedef struct Vnode{    PtrToAdjVNode FirstEdge;} AdjList[MaxVertexNum];typedef struct GNode *PtrToGNode;struct GNode{      int Nv;    int Ne;    AdjList G;};typedef PtrToGNode LGraph;LGraph ReadG(); /* details omitted */int CountConnectedComponents( LGraph Graph );int main(){    LGraph G = ReadG();    printf("%d\n", CountConnectedComponents(G));    return 0;}/* Your function will be put here */

Sample Input (for the graph shown in the figure):

8 60 70 12 04 12 43 5

Sample Output:

3code： 
int CountConnectedComponents( LGraph Graph ){    int counts = 0;//统计连通分支的个数    int i,j;    int vis[MaxVertexNum];//只标记这个点是否已经看过了    for(i = 0; i < MaxVertexNum; i++)           vis[i] = 0;//标记数组的初始化    //用邻接表实现BFS    for(i = 0; i < Graph->Nv; i++){        if(vis[i]==0){            counts++;            vis[i] = 1;            int q[MaxVertexNum];            int tail = 0,head = 0;            q[tail++] = i;//创建一个队列，并把第一个顶点入队            PtrToAdjVNode t;//创建一个指针，用于遍历这个点的邻接表            while(head<tail){//广搜                t = Graph->G[q[head]].FirstEdge;//取出队首顶点的邻接表指针，实际上指向的是首顶点下一个连接的顶点                while(t){                    if(!vis[t->AdjV]){//如果下一个连接的点没有遍历过                        vis[t->AdjV] = 1;//标记                        q[tail++] = t->AdjV;//入队                    }                    t = t->Next;//下一个连接的顶点                }                head++;//队首pop            }        }    }    return counts;}