技術標籤：PTA演算法

6-14Count Connected Components(20point(s))

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

Format of functions:

int CountConnectedComponents( LGraph Graph );

whereLGraphis 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 functionCountConnectedComponentsis 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 6
0 7
0 1
2 0
4 1
2 4
3 5

Sample Output:

3

static int visited[MaxVertexNum];
void DFS( LGraph Graph, Vertex v)
{
    PtrToAdjVNode node;
    if(visited[v]) return ;
    visited[v] = true;
    node = Graph->G[v].FirstEdge;
    while(node) {
        DFS(Graph, node->AdjV);
        node = node->Next;
    }
}
int CountConnectedComponents( LGraph Graph )
{
    int count = 0;
    for(int i = 0; i < Graph->Nv; i++) visited[i] = false;
    for(int i = 0; i < Graph->Nv; i++) {
        if(!visited[i]) {
            DFS(Graph, i);
            count++;
        }
    }
    return count;
}
 

思路：

圖的深度遍歷，每一趟訪問一個連通分量，趟數就是結果，時間複雜度是O(V+E)。