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 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
4 
1 1 
1 2 
2 3 
2 4 
3 
1 2 
2 3 
3 1

 

 

 

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。

 
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int map[1005][1005];
 6 int dgree[1005];
 7 int vis[1005];
 8 int n;
 9 int f;
10 void dfs(int x)
11 {
12     vis[x]=1;
13     for(int i=1;i<=n;i++)
14     {
15         if(!vis[i] && map[x][i])
16         {
17             dfs(i);
18             vis[i]=1;
19         }
20     }
21 }
22 void dfs1(int u,int num){
23     if(num==n){
24         f=1;
25         return;
26     }
27     for(int i=1;i<=n;i++){
28         if(!vis[i] && map[u][i]){
29             vis[i]=1;
30             dfs1(i,num+1);
31             if(f)
32                return;
33             vis[i]=0;
34         }
35     }
36 }
37 int main()
38 {
39     int t;
40     
41     int x,y;
42     while(scanf("%d",&n)==1)
43     {
44         memset(vis,0,sizeof(vis));
45         memset(map,0,sizeof(map));
46         memset(dgree,0,sizeof(dgree));
47         for(int i=0;i<n;i++)
48         {
49             scanf("%d%d",&x,&y);
50             map[x][y]=map[y][x]=1;
51             ++dgree[x];
52             ++dgree[y];
53         }
54         dfs(1);
55         int flag=1;
56         int count=0;
57         for(int i=1;i<=n;i++)
58         {
59             if(!vis[i])
60             {
61                 flag=0;
62                 break;
63             }
64             if(dgree[i]==1)
65             {
66                 count++;
67             }
68         }
69         if(flag==0 || count>2)
70         {
71             printf("NO\n");
72             continue;
73         }
74         if(count==0)
75         {
76             printf("YES\n");
77         }
78         else {
79             f=0;
80             for(int i=1;i<=n;i++){
81                 if(dgree[i]==1){
82                     memset(vis,0,sizeof(vis));
83                     vis[i]=1;
84                     dfs1(i,1);
85                     if(f)
86                       break;
87                 }
88             }
89             if(f) puts("YES");
90             else puts("NO");
91         }
92         
93     }
94     return 0;
95 }
View Code

 

 
posted @ 2015-08-29 23:50  UniqueColor  阅读(316)  评论(0)  编辑  收藏  举报