pat05-图1. List Components (25)

05-图1. List Components (25)

时间限制
200 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue

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 }

提交代码

 

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<string>
 8 using namespace std;
 9 bool map[15][15];
10 bool vis[15];
11 queue<int> q;
12 int n,e;
13 void DFS(int a){
14     q.push(a);
15     vis[a]=true;
16     int i;
17     for(i=0;i<n;i++){
18         if(!vis[i]&&map[a][i]){
19             DFS(i);
20         }
21     }
22 }
23 void BFS(int a){
24     queue<int> qq;
25     qq.push(a);
26     q.push(a);
27     vis[a]=true;
28     int i;
29     while(!qq.empty()){
30         a=qq.front();
31         qq.pop();
32         for(i=0;i<n;i++){
33             if(!vis[i]&&map[a][i]){
34                 vis[i]=true;
35                 q.push(i);
36                 qq.push(i);
37             }
38         }
39     }
40 }
41 int main(){
42     //freopen("D:\\INPUT.txt","r",stdin);
43     int a,b;
44     int i;
45     scanf("%d %d",&n,&e);
46     memset(map,false,sizeof(map));
47     memset(vis,false,sizeof(vis));
48     for(i=0;i<e;i++){
49         scanf("%d %d",&a,&b);
50         map[a][b]=map[b][a]=true;
51     }
52     for(i=0;i<n;i++){
53         if(!vis[i]){
54             DFS(i);
55             if(!q.empty()){
56                 printf("{");//{ 0 1 4 2 7 }
57                 while(!q.empty()){
58                     printf(" %d",q.front());
59                     q.pop();
60                 }
61                 printf(" }\n");
62             }
63         }
64     }
65     memset(vis,false,sizeof(vis));
66     for(i=0;i<n;i++){
67         if(!vis[i]){
68             BFS(i);
69             if(!q.empty()){
70                 printf("{");//{ 0 1 4 2 7 }
71                 while(!q.empty()){
72                     printf(" %d",q.front());
73                     q.pop();
74                 }
75                 printf(" }\n");
76             }
77         }
78     }
79     return 0;
80 }

 

posted @ 2015-08-21 16:27  Deribs4  阅读(220)  评论(0)  编辑  收藏  举报