hust1017Exact cover【dancing links 模板题】

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

Source

dupeng

研究了一上午舞蹈链是什么玩意，感觉就是把链表的便于插入删除的优点发挥到极致。

先说这货解决的是什么问题吧，精确覆盖以及延伸出来的数独求解。精确覆盖是指在一个0、1矩阵中只保留某些行，使得剩下的矩阵中每列只有一个“1“，很容易想到深搜，但是，矩阵是不断变化规模的，常见的方法不能应对，神奇的十字循环双向链表出现了，从字面就知道，每个点都有”指针“指向上下左右的点，当然了，点都是表序号的。循环体现在某行最左边的点的左域是最右边的，其他三种同理。8个数组的作用：U[],D[],R[],L[]没得说，上下左右；row[] col[]分别表示行标和列标；h[]用来进行新加入某个数字的过渡数组；s[]表示某一列元素的个数，用以剪枝优化深搜过程

模板题，不是自己代码

/************************************************
hust1017
2016,3,1

************************************************/

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxnode = 100010;
const int MaxM = 1010;
const int MaxN = 1010;
struct DLX
{
    int n,m,size;
    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],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;
        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;
    }
    //d为递归深度
    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] < S[c])//只是优化
                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 g;
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,m;
    while(scanf("%d%d",&n,&m) == 2)
    {
        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;
}