HUST 1017 Exact cover

Exact cover

http://acm.hust.edu.cn/problem.php?id=1017

Special Judge Time Limit: 15 Sec  Memory Limit: 128 MB Submissions: 5179  Solved: 2748

Description

There is an N*M matrix with only 0s and 1s, (1 <= N,M <= 1000). An exact cover is a selection of rows such that every column has a 1 in exactly one of the selected rows. Try to find out the selected rows.

Input

There are multiply test cases. First line: two integers N, M; The following N lines: Every line first comes an integer C(1 <= C <= 100), represents the number of 1s in this row, then comes C integers: the index of the columns whose value is 1 in this row.

Output

First output the number of rows in the selection, then output the index of the selected rows. If there are multiply selections, you should just output any of them. If there are no selection, just output "NO".

Sample Input

6 7
3 1 4 7
2 1 4
3 4 5 7
3 3 5 6
4 2 3 6 7
2 2 7

Sample Output

3 2 4 6

HINT

 

Source

dupeng

 

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 
  5 using namespace std;
  6 
  7 const int N=1010;
  8 const int V=1000010;
  9 
 10 int U[V],D[V];
 11 int L[V],R[V];
 12 int C[V];
 13 int H[N],S[N],mark[V];
 14 int size,n,m,vis[N],flag;
 15 
 16 void Link(int r,int c){
 17     S[c]++;C[size]=c;
 18     U[size]=U[c];D[U[c]]=size;
 19     D[size]=c;U[c]=size;
 20     if(H[r]==-1)
 21         H[r]=L[size]=R[size]=size;
 22     else{
 23         L[size]=L[H[r]]; R[L[H[r]]]=size;
 24         R[size]=H[r];  L[H[r]]=size;
 25     }
 26     mark[size]=r;
 27     size++;
 28 }
 29 
 30 void remove(int c){ //删除列
 31     L[R[c]]=L[c];
 32     R[L[c]]=R[c];
 33     int i,j;
 34     for(i=D[c];i!=c;i=D[i])
 35         for(j=R[i];j!=i;j=R[j]){
 36             U[D[j]]=U[j],D[U[j]]=D[j];
 37             S[C[j]]--;
 38         }
 39 }
 40 
 41 void resume(int c){
 42     int i,j;
 43     for(i=U[c];i!=c;i=U[i])
 44         for(j=L[i];j!=i;j=L[j]){
 45             U[D[j]]=j,D[U[j]]=j;
 46             S[C[j]]++;
 47         }
 48     L[R[c]]=c;
 49     R[L[c]]=c;
 50 }
 51 
 52 void Dance(int k){
 53     int i,j,min,c;
 54     if(!R[0]){
 55         flag=1;
 56         printf("%d",k);
 57         for(i=0;i<k;i++)
 58             printf(" %d",mark[vis[i]]);
 59         printf("\n");
 60         return ;
 61     }
 62     for(min=N,i=R[0];i;i=R[i])
 63         if(S[i]<min){
 64             min=S[i];
 65             c=i;
 66         }
 67     remove(c);
 68     for(i=D[c];i!=c;i=D[i]){
 69         vis[k]=i;
 70         for(j=R[i];j!=i;j=R[j])
 71             remove(C[j]);
 72         Dance(k+1);
 73         if(flag)
 74             return ;
 75         for(j=L[i];j!=i;j=L[j])
 76             resume(C[j]);
 77 
 78     }
 79     resume(c);
 80 }
 81 
 82 int main(){
 83 
 84     freopen("input.txt","r",stdin);
 85 
 86     while(~scanf("%d%d",&n,&m)){
 87         int i;
 88         for(i=0;i<=m;i++){
 89             S[i]=0;
 90             D[i]=U[i]=i;
 91             L[i+1]=i;R[i]=i+1;
 92         }
 93         R[m]=0;
 94         size=m+1;
 95         memset(H,-1,sizeof(H));
 96         memset(mark,0,sizeof(mark));
 97         int k,j;
 98         for(i=1;i<=n;i++){
 99             scanf("%d",&k);
100             while(k--){
101                 scanf("%d",&j);
102                 Link(i,j);
103             }
104         }
105         flag=0;
106         Dance(0);
107         if(!flag)
108             printf("NO\n");
109     }
110     return 0;
111 }

 

posted @ 2013-04-17 16:19  Jack Ge  阅读(268)  评论(0)  编辑  收藏  举报