多校5 hdu4917

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Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 365    Accepted Submission(s): 98

Problem Description bobo has a permutation p1,p2,…,pn of 1,2,…,n.

Knowing m extra constraints of form pai<pbi, bobo wanna count the number of different permutations modulo (109+7).

It is guaranteed that there is at least one such permutation. 

 

Input The input consists of several tests. For each tests:

The first line contains n,m (1≤n≤40,0≤m≤20). Each of the following m lines contain 2 integers ai,bi(1≤ai,bi≤n). 

 

Output For each tests:

A single number denotes the number of permutations. 

 

Sample Input3 11 23 21 22 3 

 

Sample Output31 

 

AuthorXiaoxu Guo (ftiasch)   

 

 求得每個連通子圖的拓撲排序的數量，在乘上相應的組合數兩種方式進行狀壓效果都是一樣的，不過bfs感覺比較好理解。都是從已經結束的狀態轉移到新的狀態。  

  1 #include <iostream>  2 #include <cstdio>  3 #include <cstring>  4 #include <queue>  5   6 using namespace std;  7   8 typedef long long LL;  9 const LL N=43; 10 LL fa[N]; 11 LL n,m; 12 LL fin(LL x){return x==fa[x]?x:fa[x]=fin(fa[x]);} 13 LL _a[N]; 14 LL _b[N]; 15 LL d[1<<21]; 16 LL cen[N]; 17 LL C[N][N]; 18 LL vis[N]; 19 LL new_id[N]; 20 const LL mod=1e9+7; 21 LL q[1<<21+1]; 22 LL head,tail; 23 int inq[1<<21]; 24 void updata(LL &a,LL b){ 25     a=a+b; 26     if(a>=mod)a-=mod; 27 } 28 LL solve(LL nn){ 29  30     head=tail=0; 31     q[tail++]=0; 32     memset(d,0,sizeof(LL)*(1<<nn)); 33     d[0]=1; 34     inq[0]=1; 35     memset(inq,0,sizeof(int)*(1<<nn)); 36     while(head<tail){ 37         LL u=q[head++]; 38  39         for(LL i=0;i<nn;i++)if((u&(1<<i))==0&&(u|cen[i])==u){ 40             updata(d[u|(1<<i)],d[u]); 41             if(!inq[u|(1<<i)]){ 42                 q[tail++]=u|(1<<i); 43                 inq[u|(1<<i)]=1; 44             } 45         } 46     } 47     return d[(1<<nn)-1]; 48 } 49  50 /*LL solve(LL nn){ 51     memset(d,0,sizeof(LL)*(1<<nn)); 52     d[0]=1; 53     for(LL u=0;u<(1<<nn);u++){ 54         for(LL i=0;i<nn;i++)if((u&(1<<i))==0&&(u|cen[i])==u){ 55             updata(d[u|(1<<i)],d[u]); 56         } 57     } 58     return d[(1<<nn)-1]; 59 }*/ 60 void cal(){ 61     //c(n,m)=c(n-1,m-1)+c(n-1,m); 62     for (int i = 0; i < N; ++ i) { 63         C[i][0] = 1; 64         for (int j = 1; j <= i; ++ j) { 65             C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod; 66         } 67     } 68 } 69 int main() 70 { 71  72     cal(); 73     while(scanf("%I64d%I64d",&n,&m)!=EOF){ 74  75         memset(vis,0,sizeof(vis)); 76  77         for(LL i=0;i<=n;i++)fa[i]=i; 78         for(LL i=0;i<m;i++){ 79             scanf("%I64d%I64d",_a+i,_b+i); 80             --_a[i];--_b[i]; 81             if(fin(_a[i])!=fin(_b[i])){ 82                 fa[fin(_a[i])]=fin(_b[i]); 83             } 84         } 85  86         LL num=n,ret=1; 87         for(LL i=0;i<n;i++)if(!vis[i]){ 88             LL _id=0; 89             for(LL j=0;j<n;j++){ 90                 if(fin(i)==fin(j)){ 91                     vis[j]=1; 92                     new_id[j]=_id++; 93                 } 94             } 95             ret=ret*C[num][_id]%mod; 96             num-=_id; 97             memset(cen,0,sizeof(cen)); 98             for(LL j=0;j<n;j++){//預先處理每個狀態其所有祖先，像貝爾曼一樣鬆弛 99                 for(LL k=0;k<m;k++)if(fin(i)==fin(_a[k])){100                     cen[new_id[_b[k]]]|=cen[new_id[_a[k]]]|(1<<new_id[_a[k]]);101                 }102             }103             ret=ret*solve(_id)%mod;104         }105         cout<<ret<<endl;106     }107     return 0;108 }

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