c编写List Components

最新推荐文章于 2024-03-06 02:29:07 发布

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最新推荐文章于 2024-03-06 02:29:07 发布

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分类专栏： ds

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本文介绍了一个给定无向图中所有连通分量的求解方法，分别使用深度优先搜索（DFS）和广度优先搜索（BFS）两种经典算法进行实现，并提供了完整的C语言代码示例。

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List Components (10分)

For a given undirected graph with N vertices and E edges, please list all the connected components by both DFS (Depth First Search) and BFS (Breadth First Search). 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 $\le$ 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 { $v_1$ $v_2$ ... $v_k$ }. 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 }













#include<stdio.h>
int N,E;
int graph[20][20]={-1};
int visited[20]={0};
int queue[20]={-1};
void DFS(int v);
void BFS(int v,int front,int rear);
int main(void)
{
	int v1,v2;
	int i=0;
	scanf("%d%d",&N,&E);
	for(i=0;i<E;i++){
		scanf("%d%d",&v1,&v2);
		graph[v1][v2]=1;
		graph[v2][v1]=1;
	}
	for(i=0;i<N;i++){
		if(visited[i]!=0)
			continue;
		else{
			printf("{");
			DFS(i);
			printf(" }\n");
		}
	}
	
	for(i=0;i<N;i++)
		visited[i]=0;
	
	for(i=0;i<N;i++){
		if(visited[i]!=0)
			continue;
		else{
			visited[i]=1;
			queue[0]=i;
			printf("{");
			BFS(i,0,1);
			printf(" }\n");
		}
	}
	return 0;
}
void DFS(int v)
{
	int w;
	visited[v]=1;
	printf(" %d",v);
	for(w=0;w<N;w++){
		if(graph[v][w]==1){
			if(visited[w]==0)
				DFS(w);
		}
	}
}
void BFS(int v,int front,int rear)
{
    int w;
	while(1){
	for(w=0;w<N;w++){
		if(graph[v][w]==1){
			if(visited[w]==0){
				queue[rear++]=w;
				visited[w]=1;
			}
		}
	}
	printf(" %d",queue[front++]);
	if(front==rear)
		break;
	v=queue[front];
	}
	
}