Another Eight Puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 859 Accepted Submission(s): 542
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
题目分析:模拟+搜索,不用剪枝都能过,只要cnt>1跳出搜索即可
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#define MAX 11
using namespace std;
int t;
int num[MAX];
int ans[MAX];
bool used[MAX];
struct Edge
{
int v,next;
}e[MAX*10];
int head[MAX];
int cc;
void add ( int u , int v )
{
e[cc].v = v;
e[cc].next = head[u];
head[u] = cc++;
}
void init ( )
{
cc = 0;
memset ( head , -1 , sizeof ( head ) );
add ( 1 , 2 );
add ( 1 , 3 );
add ( 1 , 4 );
add ( 2 , 1 );
add ( 2 , 3 );
add ( 2 , 5 );
add ( 2 , 6 );
add ( 3 , 1 );
add ( 3 , 2 );
add ( 3 , 4 );
add ( 3 , 5 );
add ( 3 , 6 );
add ( 3 , 7 );
add ( 4 , 1 );
add ( 4 , 3 );
add ( 4 , 6 );
add ( 4 , 7 );
add ( 5 , 2 );
add ( 5 , 3 );
add ( 5 , 6 );
add ( 5 , 8 );
add ( 6 , 2 );
add ( 6 , 3 );
add ( 6 , 4 );
add ( 6 , 5 );
add ( 6 , 7 );
add ( 6 , 8 );
add ( 7 , 4 );
add ( 7 , 6 );
add ( 7 , 3 );
add ( 7 , 8 );
add ( 8 , 5 );
add ( 8 , 6 );
add ( 8 , 7 );
}
int cnt;
bool check ( int u , int digit )
{
for ( int i = head[u] ; ~i ; i = e[i].next )
{
int v = e[i].v;
if ( !num[v] ) continue;
if ( num[v] == digit-1 || num[v] == digit+1 ) return false;
}
return true;
}
void dfs ( int n )
{
if ( cnt > 1 ) return;
if ( n == 9 )
{
cnt++;
for ( int i = 1 ; i <= 8 ; i++ )
ans[i] = num[i];
return;
}
if ( num[n] )
dfs ( n+1 );
else
for ( int i = 1 ; i < 9 ; i++ )
{
if ( used[i] ) continue;
if ( check ( n , i ) )
{
num[n] = i;
used[i] = true;
dfs ( n+1 );
num[n] = 0;
used[i] = false;
}
}
}
int main ( )
{
scanf ( "%d" , &t );
int c = 1;
init ( );
while ( t-- )
{
memset ( used , 0 , sizeof ( used ) );
for ( int i = 1 ; i <= 8 ; i++ )
{
scanf ( "%d" , &num[i] );
used[num[i]] = true;
}
cnt = 0;
dfs ( 1 );
printf ( "Case %d: " , c++ );
if ( cnt > 1 ) puts ( "Not unique" );
else if ( cnt == 0 ) puts ( "No answer" );
else
{
for ( int i = 1 ; i < 8 ; i++ )
printf ( "%d " , ans[i] );
printf ( "%d\n" , ans[8] );
}
}
}