05-Figure 1. List Components (25)

05-Figure 1. List components (25) time limit MS
Memory Limit 65536 KB
Code length limit 8000 B
Procedures for the award of questions StandardAuthor Chen, Yue

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

Input Specification:

Each input file contains the one test case. For each case, the first line gives integers N (0<n<=10) and E, which is the number of vertices and the number of edges, respectively. Then E-lines follow, each described a edge by giving the both ends. All the numbers in a line is 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 60 70 12 04 12 43 5

Sample Output:

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

#include <stdio.h>void DFS (int graph[][10], int *visited, int v, int n, int *ret) {Visited[v] = 1;ret[++ret[0]] = v; for (int i = 0; i < n; ++i) {if (v! = i && graph[v][i] &&!visited[i]) {DFS (graph, visited, I, N, ret); }}}void BFS (int graph[][10], int *visited, int v, int n) {int queque[20] = {};int head = 0, rear = 0;queque[rear++] = v;// Queue Visited[v] = 1;printf ("{"), while (Rear > head) {int curr = queque[head++];//out of the queue printf ("%d", Curr); for (int i = 0; I < n; ++i)//queue each visited adjacency node if (Graph[curr][i] &&!visited[i]) {visited[i] = 1;queque[rear++] = i;}} printf ("}\n");}  int main () {//freopen ("test.txt", "R", stdin); int n, edge;scanf ("%d%d", &n, &edge); int graph[10][10] = {};for (int i = 0; I < edge; ++i) {//adjacency matrix way construct diagram int v1, v2;scanf ("%d%d", &v1, &v2); Graph[v1][v2] = 1;graph[v2][v1] = 1;}  int visited[10] = {};for (int i = 0; i < n; ++i) {int ret[11] = {};//save recursive Traversal to node if (!visited[i]) {DFS (graph, visited, I, N, ret);p rintf ("{"); for (int j = 1; J <= Ret[0]; ++J) {printf ("%d", Ret[j]);} printf ("}\n");}} for (int i = 0; i < n; ++i)//reset visited tag visited[i] = 0;for (int i = 0; i < n; ++i) {//bfsif (!visited[i]) {BFS (graph, VI Sited, I, n);}} return 0;}

Title Link: http://www.patest.cn/contests/mooc-ds/05-%E5%9B%BE1

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05-Figure 1. List Components (25)