AcWing 181. Turnaround game solution (search ida* search)

AcWing 181. Swing Game
IDA* Search for , The estimated value is f(), Calculate the number with the most eight blocks in the middle 8 It's still a few minutes away , This search does not explicitly limit the number of search layers

/* 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 */

#include<bits/stdc++.h>

using namespace std;

const int N = 24;

int q[N];
int oppsite[8] = {
    5, 4, 7, 6, 1, 0, 3, 2};  // The opposite operation corresponding to the subscript sequence number  
int center[8] = {
    6, 7, 8, 11, 12, 15, 16, 17};  // The subscript of the middle eight blocks 
int path[N]; 
int op[8][7] = {
    
    {
    0, 2, 6, 11, 15, 20, 22},
    {
    1, 3, 8, 12, 17, 21, 23},
    {
    10, 9, 8, 7, 6, 5, 4},
    {
    19, 18, 17, 16, 15, 14, 13},
    {
    23, 21, 17, 12, 8, 3, 1},
    {
    22, 20, 15, 11, 6, 2, 0},
    {
    13, 14, 15, 16, 17, 18, 19},
    {
    4, 5, 6, 7, 8, 9, 10}
};

int f(){
      // Prediction function , It takes at least a few steps to find the middle circle to be consistent  
	int state[4] = {
    0};
	for(int i = 0; i < 8; i ++ ){
    
		state[q[center[i]]] ++ ;
	}
	int maxv= 0;
	for(int i = 1; i <= 3; i ++ ){
    
		maxv = max(maxv, state[i]);
	}
	return 8 - maxv;
}

void operate(int x){
    
	int t = q[op[x][0]];
	for(int i = 0; i < 6; i ++ ){
    
		q[op[x][i]] = q[op[x][i + 1]];
	}
	q[op[x][6]] = t;
}

bool dfs(int depth, int max_depth, int last){
    
	if(f() + depth > max_depth) return false;
	if(!f()) return true;
	
	for(int i = 0; i < 8; i ++ ){
    
		if(last != oppsite[i]){
    // One of the pruning  , Avoid the reverse operation of the operation just finished  
			operate(i);
			path[depth] = i;
			if(dfs(depth + 1, max_depth, i)) return true;
			operate(oppsite[i]);
		}
	}
	return false;
}

int main()
{
    
	while(cin>>q[0], q[0]){
    
		for(int i = 1; i < 24; i ++ ) cin>>q[i];
		int depth = 0;
		while(!dfs(0, depth, -1)) depth ++ ;
		
		if(!depth) 	printf("No moves needed");
		else{
    
			for(int i = 0; i < depth; i ++ ){
    
				printf("%c", 'A' + path[i]); 
			}
		} 
		cout<<endl<<q[6]<<endl;
	}
	return 0;
}