Problem V：Another Eight Puzzle（HDU 2514）

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

    There are several test cases in each input. The first two lines of each case contain two numbers C and R. (R,C <= 25) Then R lines follow, indicating the grid. '.' stands for an empty area, and a lowercase letter stands for a pile of blocks. ('a' for one block, 'b' for two blocks, 'c' for three, and so on.)

4000
Output

    Output three lines for each case. The first two lines contains two numbers x and y, indicating the initial position of the PusherBoy. (0 <= x < R, 0 <= y < C). The third line contains a moving sequence contains 'U', 'D', 'L' and 'R'. Any correct answer will be accepted.

Sample Input

[b] 3
    7 3 1 4 5 8 0 0
    7 0 0 0 0 0 0 0[/b][b]    1 0 0 0 0 0 0 0[/b]

Sample Output

    Case 1: 7 3 1 4 5 8 6 2
    Case 2: Not unique
    Case 3: No answer
————————————————————————————————————————————题意：    给出各位的点，以及所有关联位置的点，这些关连位置上的数都不能是连续的，已经给出八个点以及各点的连接情况，两个相连的点的值不能连续，输入部分点的值，让你填其他点的值，检查能否填出已经是否有唯一解。思路：    只有八个点，将数值填入点中，检查即可。    首先把关连的点用vis表示，然后在输入时将不是零的点标记，再用dfs从搜索1~9。    对于零的位置，用判断函数找出与此位置是关联的位置，只要这些位置的值都不是连续，就可以把为零的位置赋予此值。

Source Program

#include<iostream>
#include<cstring>
using namespace std;

int a[10];
int result[10];
int vis[10];
int flag;

bool judge_num(int num,int n)//数字关联判断
{
if(n==1&&num==2)
return true;
if(n==8&&num==7)
return true;
if((n>1&&n<8)&&(num==(n-1)||num==(n+1)))
return true;
return false;
}

bool judge(int postion,int n)//对位置的判断
{
int i;

if(postion==1)//A-B,A-C,A-D的判断
{
if(judge_num(a[2],n))
return false;
if(judge_num(a[3],n))
return false;
if(judge_num(a[4],n))
return false;
return true;
}
if(postion==2)//B-A,B-C,B-E,B-F的判断
{
if(judge_num(a[1],n))
return false;
if(judge_num(a[3],n))
return false;
if(judge_num(a[5],n))
return false;
if(judge_num(a[6],n))
return false;
return true;
}
if(postion==3)//C-A,C-B,C-D,C-E,C-F,C-G的判断
{
if(judge_num(a[1],n))
return false;
if(judge_num(a[2],n))
return false;
if(judge_num(a[4],n))
return false;
if(judge_num(a[5],n))
return false;
if(judge_num(a[6],n))
return false;
if(judge_num(a[7],n))
return false;
return true;
}
if(postion==4)//D-A,D-C,D-F,D-G的判断
{
if(judge_num(a[1],n))
return false;
if(judge_num(a[3],n))
return false;
if(judge_num(a[6],n))
return false;
if(judge_num(a[7],n))
return false;
return true;
}
if(postion==5)//E-B,E-C,E-F,E-H的判断
{
if(judge_num(a[2],n))
return false;
if(judge_num(a[3],n))
return false;
if(judge_num(a[6],n))
return false;
if(judge_num(a[8],n))
return false;
return true;
}
if(postion==6)//F-B,F-C,F-D,F-E,F-G,F-H的判断
{
if(judge_num(a[2],n))
return false;
if(judge_num(a[3],n))
return false;
if(judge_num(a[4],n))
return false;
if(judge_num(a[5],n))
return false;
if(judge_num(a[7],n))
return false;
if(judge_num(a[8],n))
return false;
return true;
}
if(postion==7)//G-C,G-D,G-F,G-H的判断
{
if(judge_num(a[3],n))
return false;
if(judge_num(a[4],n))
return false;
if(judge_num(a[6],n))
return false;
if(judge_num(a[8],n))
return false;
return true;
}
if(postion==8)//H-E,H-F的判断
{
if(judge_num(a[5],n))
return false;
if(judge_num(a[6],n))
return false;
return true;
}

}

void dfs(int postion)
{
int i;

if(postion>8)
{
flag++;
for(i=1;i<=8;i++)
result[i]=a[i];
return;
}

if(a[postion]==0)//对于是0的位置进行判断
{
for(i=1;i<=8;i++)//回溯搜索
{
if(!vis[i]&&judge(postion,i))//越界判断
{
vis[i]=1;
a[postion]=i;

dfs(postion+1);

vis[i]=0;
a[postion]=0;
}
}
}
else
dfs(postion+1);
}
int main()
{
int t,i,j;

scanf("%d",&t);
for(j=1;j<=t;j++)
{
memset(vis,0,sizeof(vis));
for(i=1;i<=8;i++)
{
scanf("%d",&a[i]);
vis[a[i]]=1;
}
flag=0;

dfs(1);

if(!flag)
{
printf("Case %d: ",j);
printf("No answer\n");
}
else if(flag>1)
{
printf("Case %d: ",j);
printf("Not unique\n");
}
else
{
printf("Case %d: ",j);
printf("%d",result[1]);
for(i=2;i<=8;i++)
printf(" %d",result[i]);
printf("\n");
}
}
return 0;
}