# 1017 - Exact cover

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.

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.

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".

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


3 2 4 6

  1 /* ***********************************************
2 Author        :kuangbin
3 Created Time  :2014/5/25 22:55:25
4 File Name     :E:\2014ACM\专题学习\DLX\HUST1017.cpp
5 ************************************************ */
6
7 #include <stdio.h>
8 #include <string.h>
9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 const int maxnode = 100010;
21 const int MaxM = 1010;
22 const int MaxN = 1010;
23 struct DLX
24 {
25     int n,m,size;
26     int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
27     int H[MaxN], S[MaxM];
28     int ansd, ans[MaxN];
29     void init(int _n,int _m)
30     {
31         n = _n;
32         m = _m;
33         for(int i = 0;i <= m;i++)
34         {
35             S[i] = 0;
36             U[i] = D[i] = i;
37             L[i] = i-1;
38             R[i] = i+1;
39         }
40         R[m] = 0; L[0] = m;
41         size = m;
42         for(int i = 1;i <= n;i++)
43             H[i] = -1;
44     }
46     {
47         ++S[Col[++size]=c];
48         Row[size] = r;
49         D[size] = D[c];
50         U[D[c]] = size;
51         U[size] = c;
52         D[c] = size;
53         if(H[r] < 0)H[r] = L[size] = R[size] = size;
54         else
55         {
56             R[size] = R[H[r]];
57             L[R[H[r]]] = size;
58             L[size] = H[r];
59             R[H[r]] = size;
60         }
61     }
62     void remove(int c)
63     {
64         L[R[c]] = L[c]; R[L[c]] = R[c];
65         for(int i = D[c];i != c;i = D[i])
66             for(int j = R[i];j != i;j = R[j])
67             {
68                 U[D[j]] = U[j];
69                 D[U[j]] = D[j];
70                 --S[Col[j]];
71             }
72     }
73     void resume(int c)
74     {
75         for(int i = U[c];i != c;i = U[i])
76             for(int j = L[i];j != i;j = L[j])
77                 ++S[Col[U[D[j]]=D[U[j]]=j]];
78         L[R[c]] = R[L[c]] = c;
79     }
80     //d为递归深度
81     bool Dance(int d)
82     {
83         if(R[0] == 0)
84         {
85             ansd = d;
86             return true;
87         }
88         int c = R[0];
89         for(int i = R[0];i != 0;i = R[i])
90             if(S[i] < S[c])
91                 c = i;
92         remove(c);
93         for(int i = D[c];i != c;i = D[i])
94         {
95             ans[d] = Row[i];
96             for(int j = R[i]; j != i;j = R[j])remove(Col[j]);
97             if(Dance(d+1))return true;
98             for(int j = L[i]; j != i;j = L[j])resume(Col[j]);
99         }
100         resume(c);
101         return false;
102     }
103 };
104
105 DLX g;
106 int main()
107 {
108     //freopen("in.txt","r",stdin);
109     //freopen("out.txt","w",stdout);
110     int n,m;
111     while(scanf("%d%d",&n,&m) == 2)
112     {
113         g.init(n,m);
114         for(int i = 1;i <= n;i++)
115         {
116             int num,j;
117             scanf("%d",&num);
118             while(num--)
119             {
120                 scanf("%d",&j);
122             }
123         }
124         if(!g.Dance(0))printf("NO\n");
125         else
126         {
127             printf("%d",g.ansd);
128             for(int i = 0;i < g.ansd;i++)
129                 printf(" %d",g.ans[i]);
130             printf("\n");
131         }
132     }
133     return 0;
134 }

posted on 2014-05-26 12:55  kuangbin  阅读(2829)  评论(0编辑  收藏  举报

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