AcWing 181. 回转游戏 题解（搜索—IDA*搜索）

AcWing 181. 回转游戏
IDA*搜索，预估值就是f()，计算中间八个块最多的数字离8还差几，这次搜索没有明确限制搜索的层数

#include

using namespace std;

const int N = 24;

int q[N];
int oppsite[8] = {5, 4, 7, 6, 1, 0, 3, 2};  //与下标序号对应的相反的操作 
int center[8] = {6, 7, 8, 11, 12, 15, 16, 17};  //中间八个块的下标
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(){  //预估函数，求出中中间一圈最少还需要几步才能全部一致 
	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]){//剪枝之一 ，避免刚进行完的操作又进行他的逆操作 
			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<