[HUST 1017] Exact cover

Exact cover

Time Limit: 15s Memory Limit: 128MB

Special Judge Submissions: 6012 Solved: 3185

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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define MaxNode 100010
#define MaxN 1010
#define MaxM 1010

struct DLX
{
    int n,m,size;
    int U[MaxNode],D[MaxNode],R[MaxNode],L[MaxNode];
    int Row[MaxNode],Col[MaxNode];
    int H[MaxN],S[MaxM];
    int ansd, ans[MaxN];

    void Init(int _n,int _m)
    {
        n=_n;
        m=_m;
        for(int i=0;i<=m;i++)
        {
            S[i]=0;
            U[i]=D[i]=i;
            L[i]=i-1;
            R[i]=i+1;
        }
        R[m]=0;L[0]=m;
        size=m;
        for(int i=1;i<=n;i++)
            H[i]=-1;
    }
    void Link(int r,int c)
    {
        ++S[Col[++size]=c];
        Row[size]=r;
        U[size]=U[c];
        D[U[c]]=size;
        D[size]=c;
        U[c]=size;
        if(H[r]==-1) H[r]=L[size]=R[size]=size;
        else
        {
            L[size]=L[H[r]];
            R[L[H[r]]]=size;
            R[size]=H[r];
            L[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])
            {
                U[D[j]]=j;
                D[U[j]]=j;
                S[Col[j]]++;
            }
        }
        L[R[c]] =c;
        R[L[c]] =c;
    }
    bool Dance(int d)
    {
        if(R[0]==0)
        {
            ansd=d;
            return 1;
        }
        int c=R[0];
        for(int i=R[0];i!=0;i=R[i])
            if(S[i]<S[c]) c=i;
        Remove(c);
        for(int i=D[c];i!=c;i=D[i])
        {
            ans[d]=Row[i];
            for(int j=R[i];j!=i;j=R[j]) Remove(Col[j]);  //移除
            if(Dance(d+1)) return 1;
            for(int j=L[i];j!=i;j=L[j]) Resume(Col[j]);  //回标
        }
        Resume(c);
        return 0;
    }
}g;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        g.Init(n,m);
        for(int i=1;i<=n;i++)
        {
            int num,j;
            scanf("%d",&num);
            while(num--)
            {
                scanf("%d",&j);
                g.Link(i,j);
            }
        }
        if(!g.Dance(0)) printf("NO\n");
        else
        {
            printf("%d",g.ansd);
            for(int i=0;i<g.ansd;i++)
                printf(" %d",g.ans[i]);
            printf("\n");
        }
    }
    return 0;
}