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"); } } }转载于:https://www.cnblogs.com/Dup4/p/9433192.html
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