HUST1017(KB3-A Dancing links)

1017 - Exact cover

Time Limit: 15s Memory Limit: 128MB

Special Judge Submissions: 7270 Solved: 3754

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

精确覆盖问题，dancing links模板

//2017-03-09
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 1010;
const int M = 1010;
const int maxnode = N*M;

struct DLX
{
    int n, m, sz;
    int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];
    int H[N], S[M];
    int ansd, ans[N];

    void init(int nn, int mm)
    {
        n = nn; m = mm;
        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;
        sz = m;
        for(int i = 1; i <= n; i++)H[i] = -1;
    }

    void link(int r, int c)
    {
        ++S[Col[++sz] = c];
        Row[sz] = r;
        D[sz] = D[c];
        U[D[c]] = sz;
        U[sz] = c;
        D[c] = sz;
        if(H[r] < 0)H[r] = L[sz] = R[sz] = sz;
        else{
            R[sz] = R[H[r]];
            L[R[H[r]]] = sz;
            L[sz] = H[r];
            R[H[r]] = sz;
        }
    }

    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)
        {
            printf("%d ", d);
            for(int i = 0; i < d; i++)
                  if(i == d-1)printf("%d
", ans[i]);
                else printf("%d ", ans[i]);
            return true;
        }
        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]);
            if(Dance(d+1))return true;
            for(int j = L[i]; j != i; j = L[j])resume(Col[j]);
        }
        resume(c);
        return false;
    }
}dlx;

int main()
{
    int n, m, c, tmp;
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        dlx.init(n, m);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &c);
            for(int j = 0; j < c; j++)
            {
                scanf("%d", &tmp);
                dlx.link(i, tmp);
            }
        }
        if(!dlx.Dance(0))printf("NO
");
    }

    return 0;
}