hdu 5424 Rikka with Graph II(dfs+哈密顿路径)

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  testcases.
For each testcase, the first line contains a number n(≤n≤).
Then n lines follow. Each line contains two numbers u,v(≤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

 

Sample Output

NO 
YES

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

 

Source

BestCoder Round #53 (div.2)

 

 

给一个n条边，n个顶点的图，判定是否存在哈密顿路。

 

如果存在哈密顿路，此时路径中含有n-1条边，剩下的那一条要么是自环（这里不予考虑，因为哈密顿路必然不经过），要么连接任意两个点。不考虑自环，此时图中的点度数为1的个数必然不超过2个，有如下三种情况：

1、剩下的那条边连接起点和终点，此时所有点度数都是2，可以从任意一个顶点开始进行DFS，看能否找到哈密顿路

2、剩下的那条边连接除起点和终点外的任意两个点，此时起点和终点度数为1，任选1个开始进行DFS。

3、剩下的那条边连接起点和除终点的任意一个点，或者连接终点与除起点外的任意一个点，此时图中仅有1个点度数为1，从该点开始进行DFS即可。

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 int map[][];
 int dgree[];
 int vis[];
 int n;
 int f;
 void dfs(int x)
 {
     vis[x]=;
     for(int i=;i<=n;i++)
     {
         if(!vis[i] && map[x][i])
         {
             dfs(i);
             vis[i]=;
         }
     }
 }
 void dfs1(int u,int num){
     if(num==n){
         f=;
         return;
     }
     for(int i=;i<=n;i++){
         if(!vis[i] && map[u][i]){
             vis[i]=;
             dfs1(i,num+);
             if(f)
                return;
             vis[i]=;
         }
     }
 }
 int main()
 {
     int t;

     int x,y;
     while(scanf("%d",&n)==)
     {
         memset(vis,,sizeof(vis));
         memset(map,,sizeof(map));
         memset(dgree,,sizeof(dgree));
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&x,&y);
             map[x][y]=map[y][x]=;
             ++dgree[x];
             ++dgree[y];
         }
         dfs();
         int flag=;
         int count=;
         for(int i=;i<=n;i++)
         {
             if(!vis[i])
             {
                 flag=;
                 break;
             }
             if(dgree[i]==)
             {
                 count++;
             }
         }
         if(flag== || count>)
         {
             printf("NO\n");
             continue;
         }
         if(count==)
         {
             printf("YES\n");
         }
         else {
             f=;
             for(int i=;i<=n;i++){
                 if(dgree[i]==){
                     memset(vis,,sizeof(vis));
                     vis[i]=;
                     dfs1(i,);
                     if(f)
                       break;
                 }
             }
             if(f) puts("YES");
             else puts("NO");
         }

     }
     return ;
 }