The Domino Effect
| The Domino Effect |
A standard set of Double Six dominoes contains 28 pieces (called bones) each displaying two numbers from 0 (blank) to 6 using dice-like pips. The 28 bones, which are unique, consist of the following combinations of pips:
Bone # Pips Bone # Pips Bone # Pips Bone # Pips
1 0 | 0 8 1 | 1 15 2 | 3 22 3 | 6
2 0 | 1 9 1 | 2 16 2 | 4 23 4 | 4
3 0 | 2 10 1 | 3 17 2 | 5 24 4 | 5
4 0 | 3 11 1 | 4 18 2 | 6 25 4 | 6
5 0 | 4 12 1 | 5 19 3 | 3 26 5 | 5
6 0 | 5 13 1 | 6 20 3 | 4 27 5 | 6
7 0 | 6 14 2 | 2 21 3 | 5 28 6 | 6
All the Double Six dominoes in a set can he laid out to display a 7 x 8 grid of pips. Each layout corresponds at least one ``map" of the dominoes. A map consists of an identical 7 x 8 grid with the appropriate bone numbers substituted for the pip numbers appearing on that bone. An example of a 7 x 8 grid display of pips and a corresponding map of bone numbers is shown below.
7 x 8 grid of pips map of bone numbers
6 6 2 6 5 2 4 1 28 28 14 7 17 17 11 11
1 3 2 0 1 0 3 4 10 10 14 7 2 2 21 23
1 3 2 4 6 6 5 4 8 4 16 25 25 13 21 23
1 0 4 3 2 1 1 2 8 4 16 15 15 13 9 9
5 1 3 6 0 4 5 5 12 12 22 22 5 5 26 26
5 5 4 0 2 6 0 3 27 24 24 3 3 18 1 19
6 0 5 3 4 2 0 3 27 6 6 20 20 18 1 19
Write a program that will analyze the pattern of pips in any 7 x 8 layout of a standard set of dominoes and produce a map showing the position of all dominoes in the set. If more than one arrangement of dominoes yield the same pattern, your program should generate a map of each possible layout.
Input
The input file will contain several of problem sets. Each set consists of seven lines of eight integers from 0 through 6, representing an observed pattern of pips. Each set is corresponds to a legitimate configuration of bones (there will be at least one map possible for each problem set). There is no intervening data separating the problem sets.
Output
Correct output consists of a problem set label (beginning with Set #1) followed by an echo printing of the problem set itself. This is followed by a map label for the set and the map(s) which correspond to the problem set. (Multiple maps can be output in any order.) After all maps for a problem set have been printed, a summary line stating the number of possible maps appears.
At least three lines are skipped between the output from different problem sets while at least one line separates the labels, echo printing, and maps within the same problem set.
A sample input file of two problem sets along with the correct output are shown.
Sample Input
5 4 3 6 5 3 4 6 0 6 0 1 2 3 1 1 3 2 6 5 0 4 2 0 5 3 6 2 3 2 0 6 4 0 4 1 0 0 4 1 5 2 2 4 4 1 6 5 5 5 3 6 1 2 3 1 4 2 5 2 6 3 5 4 5 0 4 3 1 4 1 1 1 2 3 0 2 2 2 2 1 4 0 1 3 5 6 5 4 0 6 0 3 6 6 5 4 0 1 6 4 0 3 0 6 5 3 6 2 1 5 3
Sample Output
Layout #1:
5 4 3 6 5 3 4 6
0 6 0 1 2 3 1 1
3 2 6 5 0 4 2 0
5 3 6 2 3 2 0 6
4 0 4 1 0 0 4 1
5 2 2 4 4 1 6 5
5 5 3 6 1 2 3 1
Maps resulting from layout #1 are:
6 20 20 27 27 19 25 25
6 18 2 2 3 19 8 8
21 18 28 17 3 16 16 7
21 4 28 17 15 15 5 7
24 4 11 11 1 1 5 12
24 14 14 23 23 13 13 12
26 26 22 22 9 9 10 10
There are 1 solution(s) for layout #1.
Layout #2:
4 2 5 2 6 3 5 4
5 0 4 3 1 4 1 1
1 2 3 0 2 2 2 2
1 4 0 1 3 5 6 5
4 0 6 0 3 6 6 5
4 0 1 6 4 0 3 0
6 5 3 6 2 1 5 3
Maps resulting from layout #2 are:
16 16 24 18 18 20 12 11
6 6 24 10 10 20 12 11
8 15 15 3 3 17 14 14
8 5 5 2 19 17 28 26
23 1 13 2 19 7 28 26
23 1 13 25 25 7 4 4
27 27 22 22 9 9 21 21
16 16 24 18 18 20 12 11
6 6 24 10 10 20 12 11
8 15 15 3 3 17 14 14
8 5 5 2 19 17 28 26
23 1 13 2 19 7 28 26
23 1 13 25 25 7 21 4
27 27 22 22 9 9 21 4
There are 2 solution(s) for layout #2.
题意：给一副图，代表多米诺骨牌摆放方式，每两个连成一块牌，如0 0 对应1号排 0 1 对应2号排，问图可以代表几种摆放方式。
思路：dfs，每个位置的牌不是竖就是横，枚举2个方向，最多枚举28块,O(2^28),加个剪枝，如果进入枚举下一行了，当前行还有没填上的，就直接回溯。
代码：
#include <stdio.h>
#include <string.h>
const int d[2][2] = {{1, 0}, {0, 1}};
int cas = 0, g[7][8], ans, num[7][7], vis[7][8], v[30], res[7][8];
void init() {
ans = 0;
memset(v, 0, sizeof(v));
memset(vis, 0, sizeof(vis));
for (int i = 0; i < 7; i++)
for (int j = 0; j < 8; j++) {
if (i == 0 && j == 0) continue;
scanf("%d", &g[i][j]);
}
}
bool judge(int x) {
for (int j = 0; j < 8; j++)
if (vis[x][j] == 0) return true;
return false;
}
void dfs(int x, int y, int nu) {
if (nu == 28) {
ans++;
for (int i = 0; i < 7; i++) {
for (int j = 0; j < 8; j++)
printf("%4d", res[i][j]);
printf("\n");
}
printf("\n");
return;
}
if (x == 7) return;
if (y == 8) {
if (judge(x)) return;
dfs(x + 1, 0, nu);
return;
}
if (vis[x][y]) {
dfs(x, y + 1, nu);
return;
}
for (int i = 0; i < 2; i++) {
if (i == 0 && x == 6) continue;
if (i == 1 && y == 7) continue;
int xx = x + d[i][0];
int yy = y + d[i][1];
if (vis[xx][yy]) continue;
if (v[num[g[x][y]][g[xx][yy]]]) continue;
res[x][y] = res[xx][yy] = num[g[x][y]][g[xx][yy]];
vis[x][y] = vis[xx][yy] = v[num[g[x][y]][g[xx][yy]]] = 1;
dfs(x, y + 1, nu + 1);
vis[x][y] = vis[xx][yy] = v[num[g[x][y]][g[xx][yy]]] = 0;
}
}
void solve() {
if (cas) printf("\n\n\n");
printf("Layout #%d:\n\n", ++cas);
for (int i = 0; i < 7; i++) {
for (int j = 0; j < 8; j++)
printf("%4d", g[i][j]);
printf("\n");
}
printf("\nMaps resulting from layout #%d are:\n\n", cas);
dfs(0, 0, 0);
printf("There are %d solution(s) for layout #%d.\n", ans, cas);
}
void table() {
int cnt = 1;
for (int i = 0; i <= 6; i++)
for (int j = i; j <= 6; j++) {
num[i][j] = num[j][i] = cnt++;
}
}
int main() {
table();
while (~scanf("%d", &g[0][0])) {
init();
solve();
}
return 0;
}