HDU 2514 Another Eight Puzzle

Another Eight Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 753    Accepted Submission(s): 464

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

 

思路：好一个DFS

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int map[10];
int flag;
int hash[10];
int t;
int sum;
int hhhh[10];
bool yes_can(int p,int x)
{
    int pow = 0;
    if(p == 1)
        return true;
    if(p == 2)
    {
        if(map[1] != x + 1 && map[1] != x - 1)
            return true;
    }
    if(p == 3)
    {
        if(map[1] != x + 1 && map[1] != x - 1
        && map[2] != x + 1 && map[2] != x - 1)
              return true;
    }
    if(p == 4)
    {
        if(map[1] != x + 1 && map[1] != x - 1
        && map[3] != x + 1 && map[3] != x - 1)
              return true;
    }
    if(p == 5)
    {
        if(map[2] != x + 1 && map[2] != x - 1
        && map[3] != x + 1 && map[3] != x - 1)
              return true;
    }
    if(p == 6)
    {
        if(map[2] != x + 1 && map[2] != x - 1
        && map[3] != x + 1 && map[3] != x - 1
        && map[4] != x + 1 && map[4] != x - 1
        && map[5] != x + 1 && map[5] != x - 1)
               return true;
    }
    if(p == 7)
    {
        if(map[3] != x + 1 && map[3] != x - 1
        && map[4] != x + 1 && map[4] != x - 1
        && map[6] != x + 1 && map[6] != x - 1)
              return true;
    }
    if(p == 8)
    {
        if(map[5] != x + 1 && map[5] != x - 1
        && map[6] != x + 1 && map[6] != x - 1
        && map[7] != x + 1 && map[7] != x - 1)
              return true;
    }
    return false;
}
void DFS(int now)
{
    //printf("%d           %d\n",now,sum);
    if(now == sum)
    {
        flag ++;
        return ;
    }
    if(flag == 2)
    {
        return ;
    }
    for(int i = 1;i <= 8;i ++)
    {
        if(hash[i] == 0 && yes_can(hhhh[now + 1],i))
        {
            hash[i] = 1;
            map[hhhh[now + 1]] = i;
            DFS(now + 1);
            hash[i] = 0;
            if(flag == 2)
                return ;
        }
    }
}
int main()
{
    scanf("%d",&t);
    for(int l = 1;l <= t;l ++)
    {
        sum = 1;
        memset(hhhh,0,sizeof(hhhh));
        memset(hash,0,sizeof(hash));
        memset(map,0,sizeof(map));
        for(int i = 1;i <= 8;i ++)
        {
           scanf("%d",&map[i]);
           if(map[i] == 0)
           {
               hhhh[sum ++] = i;
           }
           if(map[i] != 0)
               hash[map[i]] = 1;
        }
        sum --;
        flag = 0;
        DFS(0);
        printf("Case %d:",l);
        if(flag == 0)
            printf(" No answer\n");
        if(flag == 1)
        {
            for(int i = 1;i <= 8;i ++)
                printf(" %d",map[i]);
            printf("\n");
        }
        if(flag == 2)
            printf(" Not unique\n");
    }
    return 0;
}

Source

ECJTU 2008 Autumn Contest

 

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