题意:判断是否为哈密顿图
分析:首先一种情况是不合法的:也就是度数为1的点超过2个;合法的有:,那么从度数为1的点开始深搜,如果存在一种走法能够走完n个点那么存在哈密顿路
收获:学习资料
代码:
/************************************************ * Author :Running_Time * Created Time :2015-8-29 20:37:34 * File Name :C.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e3 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; vector<int> G[N]; bool vis[N]; int n; bool DFS(int u, int dep) { if (dep == n) return true; for (int i=0; i<G[u].size (); ++i) { int v = G[u][i]; if (vis[v]) continue; vis[v] = true; if (DFS (v, dep + 1)) return true; vis[v] = false; } return false; } int main(void) { while (scanf ("%d", &n) == 1) { for (int i=1; i<=n; ++i) G[i].clear (); for (int u, v, i=1; i<=n; ++i) { scanf ("%d%d", &u, &v); G[u].push_back (v); G[v].push_back (u); } bool flag = true; int s = 0, cnt = 0; for (int i=1; i<=n; ++i) { if (G[i].size () == 1) { s = i; cnt++; } } if (cnt > 2) { puts ("NO"); continue; } if (cnt == 0) s = 1; memset (vis, false, sizeof (vis)); vis[s] = true; if (!DFS (s, 1)) flag = false; puts (flag ? "YES" : "NO"); } return 0; }