1017 - Exact cover

时间限制：15秒 内存限制：128兆

自定评测 5584 次提交 2975 次通过

题目描述

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.

输入

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.

输出

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".

样例输入

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

样例输出

3 2 4 6

 Dancing links打卡题

#include   
#include   
#include   
//精确覆盖问题的定义：给定一个由0-1组成的矩阵，是否能找到一个行的集合，使得集合中每一列都恰好包含一个1
const int MN = 1005;//最大行数
const int MM = 1005;//最大列数
const int MNN = 1e5 + 5 + MM; //最大点数  

struct DLX
{
    int n, m, si;//n行数m列数si目前有的节点数  
    //十字链表组成部分  
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    //第i个结点的U向上指针D下L左R右，所在位置Row行Col列  
    int H[MN], S[MM]; //记录行的选择情况和列的覆盖情况  
    int ansd, ans[MN];
    void init(int _n, int _m)  //初始化空表  
    {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) //初始化第一横行（表头）  
        {
            S[i] = 0;
            U[i] = D[i] = i;      //目前纵向的链是空的  
            L[i] = i - 1;
            R[i] = i + 1;         //横向的连起来  
        }
        R[m] = 0; L[0] = m;
        
        
        si = m;                 //目前用了前0~m个结点  
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }
    void link(int r, int c)    //插入点(r,c)  
    {
        ++S[Col[++si] = c];     //si++;Col[si]=c;S[c]++;  
        Row[si] = r;//si该结点的行数为r
        D[si] = D[c];//向下指向c的下面的第一个结点
        U[D[c]] = si;//c的下面的第一个结点的上面为si
        U[si] = c;//si的上面为列指针
        D[c] = si;//列指针指向的第一个该列中的元素设为si
        if (H[r]<0)//如果第r行没有元素
            H[r] = L[si] = R[si] = si;
        else
        {
            R[si] = R[H[r]];//si的右边为行指针所指的右边第一个元素
            L[R[H[r]]] = si;//行指针所指的右边第一个元素的左侧为si
            L[si] = H[r];//si的左侧为行指针
            R[H[r]] = si;//行指针的右侧为si
        }
    }
    void remove(int c)        //列表中删掉c列  
    {
        L[R[c]] = L[c];//表头操作  //c列头指针的右边的元素的左侧指向c列头指针左边的元素
        R[L[c]] = R[c];//c列头指针的左边的元素的右侧指向c列头指针右边的元素
        for (int i = D[c]; i != c; i = D[i])//遍历该列的所有元素
            for (int j = R[i]; j != i; j = R[j])
            {//对于该列的某个元素所在的行进行遍历
                U[D[j]] = U[j];//把该元素从其所在列中除去
                D[U[j]] = D[j];
                --S[Col[j]];//该元素所在的列数目减一
            }
    }
    void resume(int c)        //恢复c列  
    {
        for (int i = U[c]; i != c; i = U[i])//枚举该列元素
            for (int j = L[i]; j != i; j = L[j])//枚举该列元素所在的行
                ++S[Col[U[D[j]] = D[U[j]] = j]];//D[U[j]]=j;U[D[j]]=j;S[Col[j]]++;
        L[R[c]] = R[L[c]] = c;//c列头指针左右相连
    }
    bool dance(int d) //选取了d行  
    {
        if (R[0] == 0)//全部覆盖了  
        {
            //全覆盖了之后的操作  
            ansd = d;
            return 1;
        }
        int c = R[0];//表头结点指向的第一个列
        for (int i = R[0]; i != 0; i = R[i])//枚举列头指针
            if (S[i]