1017 - Exact cover
時間限制：15秒 內存限制：128兆
自定評測 5584 次提交 2975 次通過
- 題目描述
- 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.
- 輸入
- 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.
- 輸出
- 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".
- 樣例輸入
-
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
- 樣例輸出
-
3 2 4 6
題目鏈接：http://acm.hust.edu.cn/problem/show/1017
精確覆蓋入門題。
Dancing Links 就是一種加快搜索速度的方法，采用四向鏈表。
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2014/5/25 22:55:25 4 File Name :E:\2014ACM\專題學習\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 } 45 void Link(int r,int c) 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為遞歸深度 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); 121 g.Link(i,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 }