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;
}