A Simple Task CodeForces, taskcodeforces

A Simple Task CodeForces, taskcodeforces

A Simple Task CodeForces-11D

Question: number of simple loops that output an undirected graph. A simple ring refers to a ring without duplicate edges. Make sure that the graph has no secondary auto-ring.

Ans [I] [j] indicates the number of path entries that contain vertices in I, starting from the first vertex in I and ending with j.

For an I, enumerating the current end j (obviously not the first vertex) produces a state. Then enumerate the last end k. If it can be transferred, it will be transferred.

If the number of points in I is greater than 2 and the first point in j to I has a path, the ring is generated. At last, each ring is recorded twice and divided by 2.

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 typedef long long LL; 6 LL n,m; 7 bool ok[30][30]; 8 LL ans[1000000][22]; 9 LL anss;10 int main()11 {12     LL a,b,i,j,k,fi,p,pp;13     scanf("%lld%lld",&n,&m);14     for(i=1;i<=m;i++)15     {16         scanf("%lld%lld",&a,&b);17         ok[a][b]=ok[b][a]=true;18     }19     for(i=1;i<(1<<n);i++)20     {21         pp=__builtin_popcountll(i);22         if(pp==1)23         {24             //for(j=1;j<=n;j++)25                 ans[i][__builtin_ffsll(i)]=1;26         }27         else28         {29             fi=__builtin_ffsll(i);30             for(j=1;j<=n;j++)31                 if((i&(1<<(j-1)))&&j!=fi)32                 {33                     p=i^(1<<(j-1));34                     for(k=1;k<=n;k++)35                         if((p&(1<<(k-1)))&&ok[k][j])36                         {37                             ans[i][j]+=ans[p][k];38                         }39                     if(ok[j][fi]&&pp>2)    anss+=ans[i][j];40                 }41             42         }43     }44     printf("%lld",anss/2);45     return 0;46 }47 /*48 http://blog.csdn.net/fangzhenpeng/article/details/4907823349 http://blog.csdn.net/tobewhatyouwanttobe/article/details/3803612950 http://blog.csdn.net/kk303/article/details/962193351 http://blog.csdn.net/dreamon3/article/details/5134715152 */

Slightly improved

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 typedef long long LL; 6 LL n,m; 7 bool ok[30][30]; 8 LL ans[1000000][22]; 9 LL anss;10 int main()11 {12     LL a,b,i,j,k,fi,p,pp;13     scanf("%lld%lld",&n,&m);14     for(i=1;i<=m;i++)15     {16         scanf("%lld%lld",&a,&b);17         ok[a][b]=ok[b][a]=true;18     }19     for(i=1;i<(1<<n);i++)20     {21         pp=__builtin_popcountll(i);22         fi=__builtin_ffsll(i);23         if(pp==1)24             ans[i][fi]=1;25         else26         {27             for(j=fi+1;j<=n;j++)28                 if((i&(1<<(j-1))))29                 {30                     p=i^(1<<(j-1));31                     for(k=1;k<=n;k++)32                         if((p&(1<<(k-1)))&&ok[k][j])33                             ans[i][j]+=ans[p][k];34                     if(ok[j][fi]&&pp>2)    anss+=ans[i][j];35                 }36             37         }38     }39     printf("%lld",anss/2);40     return 0;41 }42 /*43 http://blog.csdn.net/fangzhenpeng/article/details/4907823344 http://blog.csdn.net/tobewhatyouwanttobe/article/details/3803612945 http://blog.csdn.net/kk303/article/details/962193346 http://blog.csdn.net/dreamon3/article/details/5134715147 */