hdu 5424 Rikka with Graph II 搜索

Rikka with Graph II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 911    Accepted Submission(s): 224

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has a non-direct graph with n vertices and n edges. Now he wants you to tell him if there exist a Hamiltonian path.

It is too difficult for Rikka. Can you help her?

 

Input

There are no more than 100 testcases. 

For each testcase, the first line contains a number n(1≤n≤1000).

Then n lines follow. Each line contains two numbers u,v(1≤u,v≤n) , which means there is an edge between u and v.

 

Output

For each testcase, if there exist a Hamiltonian path print "YES" , otherwise print "NO".

 

Sample Input

41 11 22 32 431 22 33 1

 

Sample Output

NOYESHintFor the second testcase, One of the path is 1->2->3If you doesn't know what is Hamiltonian path, click here (https://en.wikipedia.org/wiki/Hamiltonian_path).

 

Source

BestCoder Round #53 (div.2)

删除自环，如果自环个数>1无解，=1，判定是否联通即可。

=0，枚举每条边，删除，然后判定联通。

1号点用于判定联通，因为1可能不是路径的起点，所以要从两个方向搜索。

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<vector>using namespace std;#define maxn 3001int size[maxn];int nv[maxn],nextt[maxn],id[maxn];int head[maxn];int check[maxn];void dfs(int u,int f,int xid){    if(check[u]) return ;    //cout<<u<<"->";    check[u] = 1;    size[u] = 1;    int t= 0;    for(int i = head[u];i != -1; i = nextt[i]){        if(id[i] == xid) continue;        int v = nv[i];        if(check[v]) continue;        dfs(v,u,xid);        size[u]+=size[v];        if(u == 1 && t == 0){            t++;            continue;        }            return ;    }}int in[maxn];int main(){    int n,u,v;    while(scanf("%d",&n)!=EOF){        memset(head,-1,sizeof(head));        memset(in,0,sizeof(in));        int cnt = 0,cycle=0;        for(int i = 0;i < n; i++){            scanf("%d%d",&u,&v);            if(u == v){                cycle++;                continue;            }            nv[cnt] = u;            nextt[cnt] = head[v];            id[cnt] = i;            head[v] = cnt++;            nv[cnt] = v;            nextt[cnt] = head[u];            id[cnt] = i;            head[u] = cnt++;            if(u != v)                in[u]++,in[v]++;        }        int flag = 0;        if(cycle == 1){            memset(check,0,sizeof(check));            dfs(1,0,-1);            if(size[1] == n) flag = 1;        }        else if(cycle == 0)            for(int i = 0;i < n && flag==0; i++){                memset(check,0,sizeof(check));                dfs(1,0,i);                //cout<<endl;                if(size[1] == n) flag = 1;            }        if(flag)cout<<"YES"<<endl;        else cout<<"NO"<<endl;    }    return 0;}