Sudoku
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10881 | Accepted: 3896 |
Description
In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,
| . | 2 | 7 | 3 | 8 | . | . | 1 | . |
| . | 1 | . | . | . | 6 | 7 | 3 | 5 |
| . | . | . | . | . | . | . | 2 | 9 |
| 3 | . | 5 | 6 | 9 | 2 | . | 8 | . |
| . | . | . | . | . | . | . | . | . |
| . | 6 | . | 1 | 7 | 4 | 5 | . | 3 |
| 6 | 4 | . | . | . | . | . | . | . |
| 9 | 5 | 1 | 8 | . | . | . | 7 | . |
| . | 8 | . | . | 6 | 5 | 3 | 4 | . |
Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.
Input
The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.
Output
For each test case, print a line representing the completed Sudoku puzzle.
Sample Input
.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534. ......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3. end
Sample Output
527389416819426735436751829375692184194538267268174593643217958951843672782965341 416837529982465371735129468571298643293746185864351297647913852359682714128574936
Source
思路：每次从选择方案最少的格子填充，把每个行，列，块可取的数通过压位存到整数中，取的时候&运算，不断取lowbit。这题要尤其注意常数，以及及时剪枝，用了几乎一天从1.6s减到0.77s。
1 #include <iostream> 2 #include <fstream> 3 #include <sstream> 4 #include <cstdlib> 5 #include <cstdio> 6 #include <cmath> 7 #include <string> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <stack> 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <list> 16 #include <iomanip> 17 #include <cctype> 18 #include <cassert> 19 #include <bitset> 20 #include <ctime> 21 22 using namespace std; 23 24 #define pau system("pause") 25 #define ll long long 26 #define pii pair<int, int> 27 #define pb push_back 28 #define mp make_pair 29 #define clr(a, x) memset(a, x, sizeof(a)) 30 31 const double pi = acos(-1.0); 32 const int INF = 0x3f3f3f3f; 33 const int MOD = 1e9 + 7; 34 const double EPS = 1e-9; 35 36 /* 37 #include <ext/pb_ds/assoc_container.hpp> 38 #include <ext/pb_ds/tree_policy.hpp> 39 40 using namespace __gnu_pbds; 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T; 42 */ 43 44 45 int mmpr[11], mmpc[11], mmpf[11]; 46 char s[88]; 47 int flag, num[555], cnt[1555]; 48 struct Node { 49 int val; 50 int r, c, f; 51 inline int get_bs() { 52 return mmpr[r] & mmpc[c] & mmpf[f]; 53 } 54 inline bool operator < (const Node &n) const { 55 return n.val || (!val && cnt[mmpr[r] & mmpc[c] & mmpf[f]] 56 < cnt[mmpr[n.r] & mmpc[n.c] & mmpf[n.f]]); 57 } 58 } node[11][11]; 59 void output() { 60 for (int i = 1; i <= 9; ++i) { 61 for (int j = 1; j <= 9; ++j) { 62 printf("%d", node[i][j].val); 63 } 64 puts(""); 65 } 66 puts(""); 67 } 68 inline bool dfs(int now) { 69 if (!now) return true; 70 int sx = 1, sy = 1, tmp = 10; 71 for (int i = 1; i <= 9; ++i) { 72 for (int j = 1; j <= 9; ++j) { 73 if (node[i][j].val) continue; 74 int val = node[i][j].get_bs(); 75 if (!val) return false; 76 if (cnt[val] < tmp) { 77 tmp = cnt[val]; 78 sx = i, sy = j; 79 } 80 } 81 } 82 int tbs = node[sx][sy].get_bs(); 83 for (int q = tbs & -tbs; tbs; tbs -= q, q = tbs & -tbs) { 84 node[sx][sy].val = num[q]; 85 mmpr[sx] ^= q, mmpc[sy] ^= q, mmpf[node[sx][sy].f] ^= q; 86 if (dfs(now - 1)) return true; 87 mmpr[sx] ^= q, mmpc[sy] ^= q, mmpf[node[sx][sy].f] ^= q; 88 } 89 node[sx][sy].val = 0; 90 return false; 91 } 92 int main() { 93 //freopen("data.in", "r", stdin); 94 //freopen("data.out", "w", stdout); 95 for (int i = 1; i <= 9; ++i) { 96 num[1 << i] = i; 97 } 98 for (int i = 0; i <= 1512; ++i) { 99 cnt[i] = __builtin_popcount(i); 100 } 101 while (~scanf(" %s", s + 1) && 'e' != s[1]) { 102 //puts(s + 1); pau; 103 for (int i = 1; i <= 9; ++i) { 104 mmpr[i] = mmpc[i] = mmpf[i] = (1 << 10) - 2; 105 } 106 int now = 0; 107 for (int i = 1; i <= 9; ++i) { 108 for (int j = 1; j <= 9; ++j) { 109 node[i][j].r = i; 110 node[i][j].c = j; 111 node[i][j].f = (i - 1) / 3 * 3 + (j - 1) / 3 + 1; 112 if ('.' == s[(i - 1) * 9 + j]) { 113 node[i][j].val = 0; 114 ++now; 115 } else { 116 node[i][j].val = s[(i - 1) * 9 + j] - '0'; 117 mmpr[i] ^= 1 << node[i][j].val; 118 mmpc[j] ^= 1 << node[i][j].val; 119 mmpf[node[i][j].f] ^= 1 << node[i][j].val; 120 } 121 } 122 } 123 dfs(now); 124 for (int i = 1; i <= 9; ++i) { 125 for (int j = 1; j <= 9; ++j) { 126 //printf("%d", node[i][j].val); 127 putchar(node[i][j].val + '0'); 128 } 129 } 130 puts(""); 131 } 132 return 0; 133 } 134 /* 135 2 136 000000000 137 000000000 138 000000000 139 000000000 140 000000000 141 000000000 142 000000000 143 000000000 144 000000000 145 146 */