Another Eight Puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 659 Accepted Submission(s): 405
Problem Description
Fill the following 8 circles with digits 1~8,with each number exactly once . Conntcted circles cannot be filled with two consecutive numbers.
There are 17 pairs of connected cicles:
A-B , A-C, A-D
B-C, B-E, B-F
C-D, C-E, C-F, C-G
D-F, D-G
E-F, E-H
F-G, F-H
G-H
Filling G with 1 and D with 2 (or G with 2 and D with 1) is illegal since G and D are connected and 1 and 2 are consecutive .However ,filling A with 8 and B with 1 is legal since 8 and 1 are not consecutive .
In this problems,some circles are already filled,your tast is to fill the remaining circles to obtain a solution (if possivle).
Input
The first line contains a single integer T(1≤T≤10),the number of test cases. Each test case is a single line containing 8 integers 0~8,the numbers in circle A~H.0 indicates an empty circle.
Output
For each test case ,print the case number and the solution in the same format as the input . if there is no solution ,print “No answer”.If there more than one solution,print “Not unique”.
Sample Input
3
7 3 1 4 5 8 0 0
7 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
Sample Output
Case 1: 7 3 1 4 5 8 6 2
Case 2: Not unique
Case 3: No answer
Source
ECJTU 2008 Autumn Contest
Recommend
lcy
http://acm.hdu.edu.cn/showproblem.php?pid=2514
輸入8個數 表示 圖中a b c d e f g h的位置的值只能為1-8的數 然後相鄰的不能是連續的數 即絕對值不能為1 有些數是0 將為0的填成1-8中未使用的數
有多少種方法 如果僅有一種 輸出它 有多種 或沒有 按樣例中那樣輸出
思路 DFS 搜
[cpp] #include<stdio.h>
#include<math.h>
#include<string.h>
int a[10],vis[10],ans[10],anscnt;
int abs(int q)
{
if(q<0) return -q;
return q;
}
int ok()
{
if(abs(a[2]-a[1])!=1&&
abs(a[3]-a[1])!=1&&
abs(a[4]-a[1])!=1&&
abs(a[2]-a[3])!=1&&
abs(a[2]-a[5])!=1&&
abs(a[2]-a[6])!=1&&
abs(a[3]-a[4])!=1&&
abs(a[3]-a[5])!=1&&
abs(a[3]-a[6])!=1&&
abs(a[3]-a[7])!=1&&
abs(a[4]-a[6])!=1&&
abs(a[4]-a[7])!=1&&
abs(a[5]-a[6])!=1&&
abs(a[5]-a[8])!=1&&
abs(a[6]-a[7])!=1&&
abs(a[6]-a[8])!=1&&
abs(a[7]-a[8])!=1
)
return 1;
else return 0;
}
void DFS(int k)
{
int i,cnt=0;
if(k==9)
//注意這裡是k==9 而不是把讓它等於8之後放在調用函數的最後面 那樣的話 最後一次的賦值就會被還原為0
{
if(ok())
{
anscnt++;
if(anscnt==1)
{
for(i=1;i<=8;i++)
ans[i]=a[i];
}
}
return;
}
if(anscnt>=2) return;
if(a[k]!=0) DFS(k+1);
else
for(i=1;i<=8;i++)
{
if(!vis[i])
{
a[k]=i;
vis[i]=1;
DFS(k+1);
a[k]=0;
vis[i]=0;
}
}
}
int main()
{
int t,cas=0;
scanf("%d",&t);
while(t--)
{
anscnt=0;
int i;
memset(vis,0,sizeof(vis));
memset(ans,0,sizeof(ans));
scanf("%d %d %d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6],&a[7],&a[8]);
for(i=1;i<=8;i++) vis[a[i]]=1;
DFS(1);
printf("Case %d: ",++cas);
if(anscnt==1)
{
for(i=1;i<8;i++) printf("%d ",ans[i]);
printf("%d\n",ans[i]);
}
else if(anscnt==0) printf("No answer\n");
else printf("Not unique\n");
}
return 0;
}
