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HUST 1017 Exact cover（DLX)

編輯：關於C++

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

Sample Output

3 2 4 6

DLX：精確覆蓋和重復覆蓋。此題是精確覆蓋。學習資料;點擊打開鏈接，看了一下午，加上bin神的模板，算是懂了。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pairpil;
const int maxnnode=100100;
const int maxn=1005 ;
const int mod = 1000000007;
struct DLX{
    int n,m,size;
    int U[maxnnode],D[maxnnode],L[maxnnode],R[maxnnode],Row[maxnnode],Col[maxnnode];
    int H[maxn],S[maxn];
    int ansd,ans[maxn];
    void init(int a,int b)
    {
        n=a;  m=b;
        REPF(i,0,m)
        {
            S[i]=0;
            U[i]=D[i]=i;
            L[i]=i-1;
            R[i]=i+1;
        }
        R[m]=0; L[0]=m;
        size=m;
        REPF(i,1,n)
           H[i]=-1;
    }
    void link(int r,int c)
    {
        ++S[Col[++size]=c];
        Row[size]=r;
        D[size]=D[c];
        U[D[c]]=size;
        U[size]=c;
        D[c]=size;
        if(H[r]<0)  H[r]=L[size]=R[size]=size;
        else
        {
            R[size]=R[H[r]];
            L[R[H[r]]]=size;
            L[size]=H[r];
            R[H[r]]=size;
        }
    }
    void remove(int c)
    {
        L[R[c]]=L[c];R[L[c]]=R[c];
        for(int i=D[c];i!=c;i=D[i])
        {
            for(int j=R[i];j!=i;j=R[j])
            {
                U[D[j]]=U[j];
                D[U[j]]=D[j];
                --S[Col[j]];
            }
        }
    }
    void resume(int c)
    {
        for(int i=U[c];i!=c;i=U[i])
        {
            for(int j=L[i];j!=i;j=L[j])
                ++S[Col[U[D[j]]=D[U[j]]=j]];
        }
        L[R[c]]=R[L[c]]=c;
    }
    bool Dance(int d)
    {
        if(R[0]==0)
        {
            ansd=d;
            return true;
        }
        int c=R[0];
        for(int i=R[0];i!=0;i=R[i])
        {
            if(S[i]