7-7 列出连通集(25 分)
给定一个有N个顶点和E条边的无向图,请用DFS和BFS分别列出其所有的连通集。假设顶点从0到N−1编号。进行搜索时,假设我们总是从编号最小的顶点出发,按编号递增的顺序访问邻接点。
输入格式:
输入第1行给出2个整数N(0
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>
#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;
const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-30;
const int INF = 0x3f3f3f3f;
const int maxn = 2e2 + 5;
const int MOD = 1e9 + 7;
int G[10][10];
int V[10];
int n, e;
void dfs(int x)
{
V[x] = 1;
printf("%d ", x);
for (int i = 0; i < n; i++)
{
if (V[i] == 0 && G[x][i] == 1)
dfs(i);
}
}
queue <int>q;
void bfs()
{
int len = q.size();
for (int i = 0; i < len; i++)
{
int num = q.front();
printf("%d ", num);
q.pop();
for (int j = 0; j < n; j++)
{
if (V[j] == 0 && G[num][j])
{
V[j] = 1;
q.push(j);
}
}
}
if (q.size())
bfs();
}
int main()
{
CLR(G);
CLR(V);
scanf("%d%d", &n, &e);
int x, y;
for (int i = 0; i < e; i++)
{
scanf("%d%d", &x, &y);
G[x][y] = 1;
G[y][x] = 1;
}
for (int i = 0; i < n; i++)
{
if (V[i] == 0)
{
printf("{ ");
dfs(i);
printf("}\n");
}
}
CLR(V);
for (int i = 0; i < n; i++)
{
if (V[i] == 0)
{
printf("{ ");
q.push(i);
V[i] = 1;
bfs();
printf("}\n");
}
}
}