C++ 明星奶牛问题 (Star Cow) 解题代码优化及解析

The given code is an implementation of a solution to the problem "明星奶牛" (Star Cow). However, the code seems to have some issues and does not produce the correct output for the given problem.\n\nHere is a revised version of the code that should solve the problem correctly:\n\ncpp\n#include <iostream>\n#include <vector>\n#include <queue>\n#include <bitset>\nusing namespace std;\n\nint n, m, ans;\nvector<int> g[10005];\nbitset<10005> vis;\n\nvoid bfs(int b) {\n queue<int> q;\n q.emplace(b);\n vis[b] = 1; // 修正：将 starting node 标记为已访问\n while (!q.empty()) {\n int f = q.front();\n q.pop();\n for (int i = 0; i < g[f].size(); i++) {\n int next = g[f][i]; // 优化：将 g[f][i] 赋值给 next 提高代码可读性\n if (!vis[next]) { // 修正：检查 next 是否已访问\n vis[next] = 1;\n q.emplace(next);\n }\n }\n }\n}\n\nint main() {\n cin >> n >> m;\n for (int i = 1; i <= m; i++) {\n int u, v;\n cin >> u >> v;\n g[u].emplace_back(v);\n }\n for (int i = 1; i <= n; i++) {\n vis.reset(); // 修正：将 vis.reset() 放在循环内部\n bfs(i);\n bool bo = true;\n for (int j = 1; j <= n; j++) {\n if (i != j && !vis[j]) { // 修正：排除 starting node\n bo = false;\n break;\n }\n }\n if (bo) ans++;\n }\n cout << ans << '\n';\n return 0;\n}\n\n\nThis revised code fixes a few issues in the original code:\n- vis[i] = 0; is changed to vis[i] = 1; in the bfs function to mark the starting node as visited.\n- The condition if (!vis[g[f][i]]) is changed to if (!vis[next]) to correctly check if a node has been visited.\n- The vis.reset(); statement is moved inside the loop to reset the visited array for each starting node.\n- The condition if (!vis[j]) is changed to if (i != j && !vis[j]) to exclude the starting node from the check.\n\nWith these changes, the code should produce the correct output for the given problem.