sevenzero liked Warcraft very much, but he haven‘t practiced it for several years after being addicted to algorithms. Now, though he is playing with computer, he nearly losed and only his hero Pit Lord left. sevenzero is angry, he decided to cheat to turn defeat into victory. So he entered "whosyourdaddy", that let Pit Lord kill any hostile unit he damages immediately. As all Warcrafters know, Pit Lord masters a skill called Cleaving Attack and he can damage neighbour units of the unit he attacks. Pit Lord can choice a position to attack to avoid killing partial neighbour units sevenzero don‘t want to kill. Because sevenzero wants to win as soon as possible, he needs to know the minimum attack times to eliminate all the enemys.
Input

There are several cases. For each case, first line contains two integer N (2 ≤ N ≤ 55) and M (0 ≤ M ≤ N*N),and N is the number of hostile units. Hostile units are numbered from 1 to N. For the subsequent M lines, each line contains two integers A and B, that means A and B are neighbor. Each unit has no more than 4 neighbor units. The input is terminated by EOF.

One line shows the minimum attack times for each case.

Sample Input

5 4
1 2
1 3
2 4
4 5
6 4
1 2
1 3
1 4
4 5

Sample Output

2
3

传说中的DLX在这里就显现出其与众不同的优势，将同伙连成DLX，然后计算最小值

 1 #include<bits/stdc++.h>
 2 #define FOR(i,p,X) for(int i=X[p];i!=p;i=X[i])
 3 #define For(i,a,b) for(int i=(a),i_end=(b);i<=i_end;++i)
 4 using namespace std;
 5 const int N=60;
 6 int n,m,ans;
 7 vector<int>G[N];
 8 struct DLX{  
 9     int L[N*N],R[N*N],U[N*N],D[N*N];
10     int C[N*N],H[N],cnt[N],vis[N],id;
11     void init(){  
12         For(i,0,n){  
13             cnt[i]=0;U[i]=D[i]=i;  
14             L[i+1]=i;R[i]=i+1;  
15         }  
16         R[n]=0;id=n+1;  
17         memset(H,-1,sizeof(H));  
18     }  
19     void Link(int r,int c){  
20         cnt[c]++;C[id]=c;  
21         U[id]=U[c];D[U[c]]=id;  
22         D[id]=c;U[c]=id;  
23         if(!~H[r]) H[r]=L[id]=R[id]=id;  
24         else{  
25             L[id]=L[H[r]];R[L[H[r]]]=id;  
26             R[id]=H[r];L[H[r]]=id;  
27         }  
28         id++;  
29     }
30     void Remove(int sz){
31         FOR(j,sz,D)L[R[j]]=L[j],R[L[j]]=R[j];  
32     }  
33     void Resume(int sz){  
34         FOR(j,sz,D)L[R[j]]=R[L[j]]=j;  
35     } 
36     int h(){  
37         int res=0;  
38         memset(vis,0,sizeof(vis));  
39         FOR(i,0,R){
40             if(vis[i])continue;  
41             ++res;  
42             FOR(j,i,D)FOR(k,j,R)
43                 vis[C[k]]=1;
44         }
45         return res;  
46     }  
47     void Dance(int k){  
48         if(k+h()>=ans)return;  
49         int pos=R[0];
50         if(!pos){  
51             if(k<ans)ans=k;  
52             return;  
53         }
54         FOR(i,0,R)if(cnt[pos]>cnt[i])pos=i; 
55         FOR(i,pos,D){
56             Remove(i);
57             FOR(j,i,R)Remove(j);  
58             Dance(k+1);  
59             FOR(j,i,R)Resume(j);  
60             Resume(i);  
61         }
62     }
63 }dlx;
64 int main(){
65     while(scanf("%d%d",&n,&m)!=-1){  
66         dlx.init();
67         For(i,0,n)G[i].clear(),G[i].push_back(i);
68         For(i,1,m){
69             int u,v;
70             scanf("%d%d",&u,&v);  
71             G[u].push_back(v);G[v].push_back(u);  
72         }
73         For(i,1,n)For(j,0,G[i].size()-1)
74             dlx.Link(i,G[i][j]);
75         ans=n;
76         dlx.Dance(0);  
77         printf("%d\n",ans);
78     }
79     return 0;  
80 }