Dance links算法

　　其实Dance links只是一种数据结构，Dance links 才是一种算法。dacing links x就是一个高效的求解该类问题的算法，而这种算法，基于交叉十字循环双向

链表。下面是双向十字链表的示意图：

下面给一个使用这个算法模板的裸题：

Exact Cover

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.

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

参考代码：

#include<stdio.h>
#include<iostream>
using namespace std;
const int maxnode = 1000 * 10 + 10;
const int maxN = 1000 + 10;
const int maxM = 1000 + 10;

struct DLX {
    int n,m, size;                    //n行数，m列数，szie节点数
    int U[maxnode], D[maxnode], L[maxnode], R[maxnode], Col[maxnode], Row[maxnode];
    int H[maxN], S[maxM];            //H[i]第i行的第一个节点，S[j]第j列中节点的个数
    int ansd, ans[maxN];            //ansd解包含的行数，ans[]解

    void init(int _n, int _m)        //初始化十字链表
    {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++)
        {
            Col[i] = i; Row[i] = 0;
            U[i] = D[i] = i;
            L[i] = i + 1; R[i] = i - 1;
            S[i] = 0;
        }
        R[0] = m; L[m] = 0;
        size = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }
    void link(int r, int c)        //在第r行、c列插入一个节点
    {　　　　　　　　　　　　　　　　　　//注意，这里向下为方向
        size++;
        Col[size] = c;
        Row[size] = r;
        S[c]++;
        D[size] = D[c];
        U[D[c]] = size;
        U[size] = c;
        D[c] = size;
        if (H[r] == -1)
            H[r] = R[size] = L[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)            //移除第c列及列上节点所在的行
    {
        L[R[c]] = L[c];
        R[L[c]] = R[c];
        for (int i = D[c]; i != c; i = D[i])
            for (int j = L[i]; j != i ; j = L[j])
            {
                D[U[j]] = D[j];
                U[D[j]] = U[j];
                --S[Col[j]];
            }
    }
    void resume(int c)        //恢复第c列及列上节点所在的行，与remove刚好相反
    {
        for(int i = U[c];i != c;i = U[i])
            for (int j = R[i]; j != i; j = R[j])
            {
                ++S[Col[j]];
                D[U[j]] = j;
                U[D[j]] = j;
            }
        L[R[c]] = c;
        R[L[c]] = c;
    }
    bool Dance(int d)        //d为搜索的深度
    {
        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])
                c = i;
        remove(c);
        for (int i = D[c]; i != c; i = D[i])    //逐个尝试
        {
            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);            //这个可有可无，写只是为了完整性
        return false;
    }
};

DLX g;
int main()
{
    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;
}

有什么不对的地方，多谢指教。