HUST 1017 Exact cover （DLX不可重复覆盖）

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题意：

N个点集，M个点，问最少几个点集可覆盖M个点

思路：
不可覆盖舞蹈链模板题

代码：

#include <vector>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const ll INF=0x3f3f3f3f3f3f3f3fll;const int maxn=1010;int L[maxn*100],R[maxn*100],U[maxn*100],D[maxn*100];//节点的上下左右四个方向的链表int C[maxn*100],H[maxn],cnt[maxn],vis[maxn],ans[maxn],Row[maxn*100];//C列H行cnt列链表中元素个数int n,m,id,len;void init(){    for(int i=0;i<=m;i++){        cnt[i]=0;U[i]=D[i]=i;        L[i+1]=i;R[i]=i+1;    }    R[m]=0;id=m+1;    memset(H,-1,sizeof(H));}void Link(int r,int c){    cnt[c]++;C[id]=c;Row[id]=r;    U[id]=U[c];D[U[c]]=id;    D[id]=c;U[c]=id;Row[id]=r;    if(H[r]==-1) H[r]=L[id]=R[id]=id;    else{        L[id]=L[H[r]];R[L[H[r]]]=id;        R[id]=H[r];L[H[r]]=id;    }    id++;}void Remove(int Size){    L[R[Size]]=L[Size];    R[L[Size]]=R[Size];    for(int i=D[Size];i!=Size;i=D[i]){        for(int j=R[i];j!=i;j=R[j]){            U[D[j]]=U[j];D[U[j]]=D[j];            cnt[C[j]]--;        }    }}void Resume(int Size){    for(int i=D[Size];i!=Size;i=D[i]){        for(int j=R[i];j!=i;j=R[j]){            U[D[j]]=j;D[U[j]]=j;            cnt[C[j]]++;        }    }    L[R[Size]]=Size;R[L[Size]]=Size;}int Dance(int k){    int pos,mm=maxn;    if(R[0]==0){        len=k;        return 1;    }    for(int i=R[0];i;i=R[i]){        if(mm>cnt[i]){            mm=cnt[i];pos=i;        }    }    Remove(pos);    for(int i=D[pos];i!=pos;i=D[i]){        ans[k]=Row[i];        for(int j=R[i];j!=i;j=R[j]) Remove(C[j]);        if(Dance(k+1)) return 1;        for(int j=L[i];j!=i;j=L[j]) Resume(C[j]);    }    Resume(pos);    return 0;}int main(){    int u,v,a;    while(scanf("%d%d",&n,&m)!=-1){        init();        for(int i=1;i<=n;i++){            scanf("%d",&a);            while(a--){                scanf("%d",&u);                Link(i,u);            }        }        int flag=Dance(0);        if(flag==0) printf("NO\n");        else{            printf("%d",len);            for(int i=0;i<len;i++) printf(" %d",ans[i]);            printf("\n");        }    }    return 0;}

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 73 1 4 72 1 43 4 5 73 3 5 64 2 3 6 72 2 7

Sample Output

3 2 4 6

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