PAT 05-1 List Components （简单DFS与BFS）

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#数据结构 #图 #DFS #BFS #c语言

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本文探讨了一道编程题的解题过程，作者最初过于复杂地思考问题，后意识到使用顺序结构而非链表能更高效解决。通过对比BFS与DFS算法在寻找图的连通分量中的应用，强调了选择合适的数据结构的重要性。实例分析了如何通过输入输出规范，利用DFS和BFS分别获取连通分量，并提供了具体代码实现。

刚一拿到这道题把他想的太复杂了

明明是长度最大为十的顺序结构就能解决的问题，竟然优先想到用链表。

BFS牵扯到一个队列的操作，在这种小规模数据里面用顺序结构好很多

题目如下：

For a given undirected graph with N vertices and E edges, please list all the connected components by both DFS and BFS. Assume that all the vertices are numbered from 0 to N-1. While searching, assume that we always start from the vertex with the smallest index, and visit its adjacent vertices in ascending order of their indices.

Input Specification:

Each input file contains one test case. For each case, the first line gives two integers N (0<N<=10) and E, which are the number of vertices and the number of edges, respectively. Then E lines follow, each described an edge by giving the two ends. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in each line a connected component in the format "{ v1 v2 ... vk }". First print the result obtained by DFS, then by BFS.

Sample Input:

Sample Output:

{ 0 1 4 2 7 }
{ 3 5 }
{ 6 }
{ 0 1 2 7 4 }
{ 3 5 }
{ 6 }

题目链接：点我点我点我

代码如下：

# include <stdio.h>
# include <stdlib.h>
void DFS(int i);
void BFS(int i);

int vertice[10][10];   
int flag[10];          
int n,e; 

              
int main()
{
    int i,j,k;
    scanf("%d%d",&n,&e);
    int a,b;
    for (i=0;i<e;i++)
    {
        scanf("%d%d",&a,&b);
        vertice[a][b] = vertice[b][a] = 1;
    }
    for (i=0;i<n;i++)
       {
        if (flag[i]==1) continue;
        else  printf("{ "),DFS(i),printf("}\n");
       }
    for(i=0;i<10;i++)
          flag[i] = 0;
    for (i=0;i<n;i++)
       {
        if (flag[i]==1) continue;
        else printf("{ "),BFS(i),printf("}\n");
       }
}


void DFS(int i)
{
    printf("%d ",i),flag[i] = 1;
    int j;
    for (j=0;j<n;j++)
        if (vertice[i][j]==1&&flag[j]==0)
          {
             DFS(j);
          }
}



void BFS(int i)
{
   int start,end,Q[100];
   start = end = 0;
   Q[end++] = i;flag[i] = 1;
   int temp,j;
   while (start<end)
   {
       temp = Q[start++];
       printf("%d ",temp);
       for (j=0;j<n;j++)
          if (vertice[temp][j]==1&&flag[j]==0)
          {
              Q[end++] = j;
              flag[j] = 1;
          }
   }
}