1017 - Exact cover

时间限制：15秒 内存限制：128兆

自定评测 6110 次提交 3226 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

提示

来源

dupeng

本人智商奇低，看了三天才学会，模板验证题。

顺便一提这是正式转入C++后第一A。

  1 #include<iostream>
  2 #include<cstdio>
  3 using    namespace    std;
  4
  5 const    int    HEAD = 0;
  6 const    int    N = 1005;
  7 int    MAP[N][N];
  8 int    U[N * N],D[N * N],L[N * N],R[N * N],H[N * N],C[N * N],ANS[N * N];
  9
 10 void    ini(int col);
 11 bool    dancing(int k);
 12 void    output(void);
 13 void    remove(int c);
 14 void    resume(int c);
 15 int    main(void)
 16 {
 17     int    n,m,num,col;
 18     int    count,front,first;
 19
 20     while(cin >> n >> m)
 21     {
 22         ini(m);
 23
 24         count = m + 1;
 25         for(int i = 1;i <= n;i ++)
 26         {
 27             cin >> num;
 28             front = first = count;
 29             while(num --)
 30             {
 31                 cin >> col;
 32
 33                 U[count] = U[col];
 34                 D[count] = col;
 35                 L[count] = front;
 36                 R[count] = first;
 37
 38                 D[U[col]] = count;
 39                 U[col] = count;
 40                 R[front] = count;
 41
 42                 H[count] = i;
 43                 C[count] = col;
 44                 front = count;
 45                 count ++;
 46             }
 47             L[first] = count - 1;
 48         }
 49         if(!dancing(1))
 50             cout << "NO" << endl;
 51     }
 52
 53     return    0;
 54 }
 55
 56 void    ini(int col)
 57 {
 58     U[HEAD] = D[HEAD] = H[HEAD] = C[HEAD] = HEAD;
 59     R[HEAD] = 1;
 60     L[HEAD] = col;
 61
 62     int    front = HEAD;
 63     for(int i = 1;i <= col;i ++)
 64     {
 65         U[i] = D[i] = i;
 66         L[i] = front;
 67         R[i] = HEAD;
 68         R[front] = i;
 69         front = i;
 70
 71         C[i] = i;
 72         H[i] = 0;
 73     }
 74 }
 75
 76 bool    dancing(int k)
 77 {
 78     int    c = R[HEAD];
 79     if(c == HEAD)
 80     {
 81         output();
 82         return    true;
 83     }
 84
 85     remove(C[c]);
 86     for(int i = D[c];i != c;i = D[i])
 87     {
 88         ANS[k] = H[i];
 89         for(int j = R[i];j != i;j = R[j])
 90             remove(C[j]);
 91         if(dancing(k + 1))
 92             return    true;
 93         for(int j = L[i];j != i;j = L[j])
 94             resume(C[j]);
 95     }
 96     resume(C[c]);
 97
 98     return    false;
 99 }
100
101 void    output(void)
102 {
103     int    i,j;
104     for(i = 1;ANS[i];i ++);
105     cout << i - 1 << " ";
106     for(j = 1;j < i - 1;j ++)
107         cout << ANS[j] << " ";
108     cout << ANS[j] << endl;
109 }
110
111 void    remove(int c)
112 {
113     R[L[c]] = R[c];
114     L[R[c]] = L[c];
115
116     for(int i = D[c];i != c;i = D[i])
117         for(int j = R[i];j != i;j = R[j])
118         {
119             D[U[j]] = D[j];
120             U[D[j]] = U[j];
121         }
122 }
123
124 void    resume(int c)
125 {
126     R[L[c]] = c;
127     L[R[c]] = c;
128
129     for(int i = U[c];i != c;i = U[i])
130         for(int j = R[i];j != i;j = R[j])
131         {
132             D[U[j]] = j;
133             U[D[j]] = j;
134         }
135 }