PTA List Components
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:
8 6 0 7 0 1 2 0 4 1 2 4 3 5
Sample Output:
{ 0 1 4 2 7 }
{ 3 5 }
{ 6 }
{ 0 1 2 7 4 }
{ 3 5 }
{ 6 }
这题比较水……就是写个图的DFS和BFS……当然DFS是遍历到这个点才标记该点已经被访问,并且已经访问过的点就不要再去访问了,不然如果图中有环的话就一直递归下去了,BFS的话是只要入队就把相应节点标记(不用管它有没有遍历到,因为只要入队肯定会遍历到),这样标记过的点就不用再入队了,从而避免了重复入队的发生。
下面是代码:
//
// main.c
// List Components
//
// Created by 余南龙 on 2016/12/6.
// Copyright © 2016年 余南龙. All rights reserved.
//
#include <stdio.h>
#include <string.h>
#define MAXV 10000
int Graph[MAXV][MAXV];
int visit[MAXV], connected[MAXV];
int N, E, top;
void DFS(int v){
int i;
connected[++top] = v;
visit[v] = ;
; i < N; i++){
== Graph[v][i]&& == visit[i]){
DFS(i);
}
}
}
void BFS(int v){
];
, j = , i;
Q[++tail] = v;
visit[Q[j]] = ;
){
connected[++top] = Q[j];
; i < N; i++){
== Graph[Q[j]][i]&& == visit[i]){
Q[++tail] = i;
visit[i] = ;
}
}
j++;
if(tail < j){
break;
}
}
}
void Init(){
int i, u, v;
scanf("%d%d", &N, &E);
; i < E; i++){
scanf("%d%d", &u, &v);
Graph[u][v] = Graph[v][u] = ;
}
}
void Output(){
int i;
printf("{ ");
; i <= top; i++){
printf("%d ", connected[i]);
}
printf("}\n");
}
int main(){
int j;
Init();
memset(visit, , MAXV * sizeof(int));
top = -;
; j < N; j++){
== visit[j]){
DFS(j);
Output();
top = -;
}
}
memset(visit, , MAXV * sizeof(int));
top = -;
; j < N; j++){
== visit[j]){
BFS(j);
Output();
top = -;
}
}
;
}
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