24h購物| | PChome| 登入
2012-12-27 18:05:29| 人氣755| 回應0 | 上一篇 | 下一篇

[UVA][DLX][舞鏈] 1309 - Sudoku

推薦 0 收藏 0 轉貼0 訂閱站台

A Sudoku grid is a 16 x 16 grid of cells grouped in sixteen 4 x 4 squares, where some cells are filled with letters from A to P (the first 16 capital letters of the English alphabet), as shown in figure 1a. The game is to fill all the empty grid cells with letters from A to P such that each letter from the grid occurs once only in the line, the column, and the 4 x 4 square it occupies. The initial content of the grid satisfies the constraints mentioned above and guarantees a unique solution.

Figure 1. Sudoku




A



C




O
I

J

A
B
P
C G F
H


D

F
I
E



P

G
E L
H



M
J





E



C

G



I

K
G A
B


E
J
D
G P

J
F



A


E


C
B

D P

O
E

F
M

D

L
K
A

C







O
I
L
H
P
C

F
A

B





G
O D


J



H
K


J



H
A
P
L


B

P

E

K

A

H

B

K

F I
C



F


C

D

H
N
















a) Sudoku grid


F P A H M J E C N L B D K O G I
O J M I A N B D P K C G F L H E
L N D K G F O I J E A H M B P C
B G C E L K H P O F I M A J D N
M F H B E L P O A C K J G N I D
C I L N K D G A H B M O P E F J
D O G P I H J M F N L E C A K B
J E K A F C N B G I D P L H O M
E B O F P M I J D G H L N K C A
N C J D H B A E K M O F I G L P
H M P L C G K F I A E N B D J O
A K I G N O D L B P J C E F M H
K D E M J I F N C H G A O P B L
G L B C D P M H E O N K J I A F
P H N O B A L K M J F I D C E G
I A F J O E C G L D P B H M N K
















b) Solution

Write a Sudoku playing program that reads data sets from a text file.

Input 

Each data set encodes a grid and contains 16 strings on 16 consecutive lines as shown in figure 2. The i-th string stands for the i-th line of the grid, is 16 characters long, and starts from the first position of the line. String characters are from the set {A,B,...,P,-}, where `-' (minus) designates empty grid cells. The data sets are separated by single empty lines and terminate with an end of file.

Output 

The program prints the solution of the input encoded grids in the same format and order as used for input.

Sample Input  

--A----C-----O-I 
-J--A-B-P-CGF-H-
--D--F-I-E----P-
-G-EL-H----M-J--
----E----C--G---
-I--K-GA-B---E-J 
D-GP--J-F----A--
-E---C-B--DP--O-
E--F-M--D--L-K-A 
-C--------O-I-L-
H-P-C--F-A--B---
---G-OD---J----H 
K---J----H-A-P-L 
--B--P--E--K--A-
-H--B--K--FI-C--
--F---C--D--H-N-

Sample Output  

FPAHMJECNLBDKOGI 
OJMIANBDPKCGFLHE 
LNDKGFOIJEAHMBPC 
BGCELKHPOFIMAJDN 
MFHBELPOACKJGNID 
CILNKDGAHBMOPEFJ 
DOGPIHJMFNLECAKB 
JEKAFCNBGIDPLHOM 
EBOFPMIJDGHLNKCA 
NCJDHBAEKMOFIGLP 
HMPLCGKFIAENBDJO 
AKIGNODLBPJCEFMH 
KDEMJIFNCHGAOPBL 
GLBCDPMHEONKJIAF 
PHNOBALKMJFIDCEG 
IAFJOECGLDPBHMNK

Dancing Links 實作數獨，至於 DLX 是什麼結構，上網查一下唄。
此外題目忘了說明， "Output a empty blank line between test cases."


#include<stdio.h>
#include<stdlib.h>
#define Maxv 100000
struct DacingLinks {
    int left, right;
    int up, down;
    int data, ch, rh;
}DL[1000001 + 3001];
int s[2001], o[2001], head, size, time;
int n, m, a, b, c, tn;
void Remove(int c) {
    DL[DL[c].right].left = DL[c].left;
    DL[DL[c].left].right = DL[c].right;
    int i, j;
    for(i = DL[c].down; i != c; i = DL[i].down) {
        for(j = DL[i].right; j != i; j = DL[j].right) {
            DL[DL[j].down].up = DL[j].up;
            DL[DL[j].up].down = DL[j].down;
            s[DL[j].ch]--;
        }
    }
}
void Resume(int c) {
    int i, j;
    for(i = DL[c].down; i != c; i = DL[i].down) {
        for(j = DL[i].left; j != i; j = DL[j].left) {
            DL[DL[j].down].up = j;
            DL[DL[j].up].down = j;
            s[DL[j].ch]++;
        }
    }
    DL[DL[c].right].left = c;
    DL[DL[c].left].right = c;
}
void Print_board(int k) {
    static int set[512], a, b;
    for(a = 0; a < k; a++) {
        set[DL[o[a]].data] = DL[o[a]].rh+1;
    }
    for(a = 0; a < n; a++) {
        for(b = 0; b < n; b++) {
            printf("%c", set[a*16+b]+'A'-1);
        }
        puts("");
    }
}
void DFS(int k) {
    if(time) return ;
    if(DL[head].right == head) {
        Print_board(k);
        time++;
        return;
    }
    int t = Maxv, c, i, j;
    for(i = DL[head].right; i != head; i = DL[i].right) {
        if(s[i] < t) {
            t = s[i], c = i;
        }
    }
    Remove(c);
    for(i = DL[c].down; i != c; i = DL[i].down) {
        o[k] = i;
        for(j = DL[i].right; j != i; j = DL[j].right)
            Remove(DL[j].ch);
        DFS(k+1);
        for(j = DL[i].left; j != i; j = DL[j].left)
            Resume(DL[j].ch);
    }
    Resume(c);
}
int new_node(int up, int down, int left, int right) {
    DL[size].up = up, DL[size].down = down;
    DL[size].left = left, DL[size].right = right;
    DL[up].down = DL[down].up = DL[left].right = DL[right].left = size;
    return size++;
}
void new_row(int n, int Row[], int rh, int set) {
    int a, r, row = -1, k;
    for(a = 0; a < n; a++) {
        r = Row[a];
        DL[size].ch = r, DL[size].data = set, s[r]++;
        if(row == -1) {
            row = new_node(DL[DL[r].ch].up, DL[r].ch, size, size);
            DL[row].rh = rh;
        }else {
            k = new_node(DL[DL[r].ch].up, DL[r].ch, DL[row].left, row);
            DL[k].rh = rh;
        }
    }
}
void init(int m) {
    size = 0;
    head = new_node(0, 0, 0, 0);
    int i;
    for(i = 1; i <= m; i++) {
        new_node(i, i, DL[head].left, head);
        DL[i].ch = i, s[i] = 0;
    }
}
main() {
    int map[20][20], Row[256], t, i;
    char g[20][20], first = 0;
    while(scanf("%s", g[0]) == 1) {
        for(i = 1; i < 16; i++)
            scanf("%s", g[i]);
        n = 4;
        tn = n, n *= n, m = n*n*4, init(m);
        for(a = 0; a < n; a++)
            for(b = 0; b < n; b++) {
                if(g[a][b] == '-')
                    map[a][b] = 0;
                else
                    map[a][b] = g[a][b]-'A'+1;
                if(map[a][b] == 0) {
                    for(c = 0; c < n; c++) {
                        t = 0;
                        Row[t++] = a*n + b + 1;
                        Row[t++] = n*n + a*n + c + 1;
                        Row[t++] = 2*n*n+ b*n + c + 1;
                        Row[t++] = 3*n*n+ (a/tn*tn + b/tn)*n + c + 1;
                        new_row(t, Row, c, a*n+b);
                        /*printf("%d ", a*9 + b + 1);/*原始牌面上的標記
                        printf("%d ", 81 + a*9 + c + 1);/*行上1~9的標記
                        printf("%d ", 162+ b*9 + c + 1);/*列上1~9的標記
                        printf("%d ", 243+ (a/3*3 + b/3)*9 + c + 1);/*方格中1~9的標記
                        puts("");*/
                    }
                } else {
                    c = map[a][b]-1, t = 0;
                    Row[t++] = a*n + b + 1;
                    Row[t++] = n*n + a*n + c + 1;
                    Row[t++] = 2*n*n+ b*n + c + 1;
                    Row[t++] = 3*n*n+ (a/tn*tn + b/tn)*n + c + 1;
                    new_row(t, Row, c, a*n+b);
                    /*printf("%d ", a*9 + b + 1);/*原始牌面上的標記
                    printf("%d ", 81 + a*9 + c + 1);/*行上1~9的標記
                    printf("%d ", 162+ b*9 + c + 1);/*列上1~9的標記
                    printf("%d ", 243+ (a/3*3 + b/3)*9 + c + 1);/*方格中1~9的標記
                    puts("");*/
                }
            }
        if(first)   puts("");
        first = 1;
        time = 0, DFS(0);

    }
    return 0;
}
 

我要檢舉
#1309#Sudoku#DLX#舞鏈
台長： Morris

您可能對以下文章有興趣

人氣(755) | 回應(0)| 推薦 (0)| 收藏 (0)| 轉寄
全站分類: 不分類 | 個人分類: UVA |
此分類下一篇:[UVA][DLX][舞鏈] 387 - A Puzzling Problem
此分類上一篇:[UVA][dp] 12589 - Learning Vector

我要回應
否 是 (若未登入"個人新聞台帳號"則看不到回覆唷!)
* 請輸入識別碼：
請輸入圖片中算式的結果(可能為0) 
(有*為必填)
TOP
詳全文