【HDU】2514 Another Eight Puzzle（C++/C）

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

 

这道题就是一般的dfs，不过有17条边，需要定义一个check函数

代码：

//
//  main.cpp
//  2514
//
//  Created by showlo on 2018/5/17.
//  Copyright © 2018年 showlo. All rights reserved.
//

#include <iostream>
#include <stdio.h>
using namespace std;

int vis[10];

int a[10],num,ans[10];

int abs(int q)            //定义求绝对值的函数
{
    if(q<0) return -q;
    return q;
}

int check(){               //检查此时的数组是否符合要求
    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 n){
    if(n==9&&check()){         //如果已经到第9个了，就可以进行检查了
        num++;
        if(num==1){
            for(int i=1;i<=8;i++)
                ans[i]=a[i];
        }
        //for(int i=1;i<=8;i++)
          //  cout<<" "<<ans[i];
        //cout<<endl;
        return;
    }
    
    if(num>=2)
        return;
    if(a[n]!=0)
        dfs(n+1);
    else
        for (int i=1; i<=8; i++) {
            if (vis[i]==0) {
                a[n]=i;
                vis[i]=1;
                dfs(n+1);
                vis[i]=0;
                a[n]=0;
            }
        }
}


int main() {
    int n,cas=0;
    cin>>n;
    while (n--) {
        num=0;
        memset(vis, 0, sizeof(vis));
        memset(ans, 0, sizeof(ans));
        for(int i=1;i<=8;i++)
            cin>>a[i];
        for (int i=1; i<=8; i++)
            vis[a[i]]=1;  //记录已经用过的数
        dfs(1);
        cout<<"Case "<<++cas<<":";
        if(num>=2)
            cout<<" Not unique"<<endl;
        else if(num==1)
        {
            for(int i=1;i<=8;i++)
                cout<<" "<<ans[i];
            cout<<endl;
        }
        else if(num==0)
            cout<<" No answer"<<endl;
    }
    return 0;
}