HUST 1017 Exact cover DLX

题意：一个n×m行的矩阵，每个格子可能是 0 或者1  现在让你选择 几行  使得 每一列 有且只有一个1.（精确覆盖模板题）

解题思路：

dancing links 模板 

关于dancing links  献上几篇必看论文 ：http://par.buaa.edu.cn/acm-icpc/filepool/r/35/

板子用的kuangbin的板子（还是适牛的。。）

解题思路：

// File Name: hust1017.cpp
// Author: darkdream
// Created Time: 2014年10月04日 星期六 16时47分57秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long
const int maxnode = 100010;
const int MaxM = 1010;
const int MaxN = 1010;
using namespace std;
struct DLX
{
    int n ,m ,size;
    int L[maxnode],R[maxnode],U[maxnode],D[maxnode],Row[maxnode],Col[maxnode];
    int ansd,ans[MaxN];
    int H[MaxN],S[MaxM];
    void init(int _n,int _m)
    {
        n = _n;
        m = _m;
        for(int i = 0;i <= m;i++)
        {
            S[i] = 0; // S 代表的是每一列的颜色
            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;  //  H[i]  代表第I 行的头
    }
    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; // 插入到  H[r] 的右边
        }
    }
    void remove(int c) //列为单位，横向
    {
        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]];
            }//删除一列，k行
    }
    void resume(int c)
    {
        L[R[c]] = R[L[c]] = 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]];
            }
    }
    bool Dance(int d)
    {
       if(R[0] == 0 )
       {
          ansd = d; //答案长度
          return true;
       }
       int c = R[0];
       for(int i = R[c];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 g;
int main(){
   int n , m ; 
   while(scanf("%d %d",&n,&m) != EOF)
   {
      g.init(n,m);
      int num, j; 
      for(int i = 1;i <= n;i ++)
      {
        scanf("%d",&num);
        while(num--)
        {
          scanf("%d",&j);
          g.Link(i,j);
        }
      }
      if(!g.Dance(0)) printf("NO
");
      else{
        printf("%d ",g.ansd);
        for(int i = 0;i < g.ansd;i ++)
          printf(" %d",g.ans[i]);
        printf("
"); 
      }
      
   }
return 0;
}