HDU5424——DFS——Rikka with Graph II

http://acm.hdu.edu.cn/showproblem.php?pid=5424

/*
对于一个连通图，有N个顶点，如果有N条边，那么肯定有环
因为N个点如果不成环最多N-1条边。。。然后找最小度的点DFS就行。。。 
*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-31 17:29:13
* File Name     :HDU5424.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 MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int deg[MAXN];
int n;
int mp[MAXN][MAXN];
int vis[MAXN];
vector<int> G[MAXN];
int flag;
void dfs(int u, int num)
{
    if(flag == 1) return;
    if(num == n){
        flag = 1;
        return ;
    }
    for(int i = 0; i < G[u].size(); i++){
        int v = G[u][i];
        if(!vis[v]){
            vis[v] = 1;
            dfs(v, num+1);
            vis[v] = 0;
        }
    }
}
int main(){
    int u, v;
    while(~scanf("%d", &n)){
        memset(mp, 0, sizeof(mp));
        memset(vis, 0, sizeof(vis));
        memset(deg, 0, sizeof(deg));
        for(int i = 1; i <= n; i++)
            G[i].clear();
        for(int i = 1; i <= n; i++){
            scanf("%d%d", &u, &v);
            if(mp[u][v] == 0 ){
            mp[u][v] = mp[v][u] = 1;
            G[u].push_back(v);
            G[v].push_back(u);
            deg[u]++;
            deg[v]++;
            }
        }
        int tot = 0;
        int id = 1;
        for(int i = 1; i <= n; i++)
            if(deg[i] == 1){
                id = i;
                tot++;
            }
        if(tot > 2){
            puts("NO");
            continue;
        }
        flag = 0;
        vis[id] = 1;
        dfs(id, 1);
        printf("%s
", !flag ? "NO" : "YES");
    }
    return 0;
}