題意：有8個立方體，按相同顏色著色，相對面是相同顏色，按同一方向擺成3*3的方陣。問從初始狀態到目標狀態最少移動步數。

分析：題目是最短路徑題，可以bfs可以回溯，用回溯加剪枝。不能走的格子是上次走的空格。

# include<iostream>
# include<cstdio>
# include<cmath>
# include<map>
# include<queue>
# include<cstring>
#include<set>
# include<algorithm>
using namespace std;

struct node {
	int top, front;
};
int goal[3][3];
node start[3][3];
int dx[] = { 1,-1,0,0 };
int dy[] = { 0,0,1,-1 };
int step = 31;
int judge() {
	int t = 0;
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 3; j++) {
			if (start[i][j].top != goal[i][j])t++;
		}
	}
	return t;
}
void move(int x,int y,int i) {
	int ddx = x + dx[i]; int ddy = y + dy[i];
	if (i == 0 || i == 1) {
		swap(start[x][y].front, start[x][y].top);
	}
	else {
		start[x][y].top = 1 ^ 2 ^ 3 ^ start[x][y].front^start[x][y].top;
	}
	swap(start[x][y], start[ddx][ddy]);
}
void dfs(int curx,int cury,int layer) {
	if (layer > 30)return;
	int jp = judge();
	if (jp==0) {
		step = min(step, layer);
		return;
	}
	if (jp + layer > step)return;//剪枝
	int tx, ty;
	for(int i=0;i<3;i++)
		for(int j=0;j<3;j++)
			if (!start[i][j].top) {
				tx = i; ty = j; break;
			}
	for (int i = 0; i < 4; i++) {
		int ddx = tx + dx[i];
		int ddy = ty + dy[i];
		if (ddx >= 0 && ddx < 3 && ddy>=0 && ddy < 3 && (ddx != curx || ddy != cury)) {
			move(ddx, ddy, i ^ 1);
			dfs(tx, ty, layer + 1);
			move(tx, ty, i);
		}
	}
}
int main() {
	int ex, ey;
	char k;
	//0-empty,1-red,2-blue,3-white
	while (cin >> ey >> ex && ex&&ey) {
		ex--; ey--;
		step = 31;
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++) {
				cin >> k;
				if (k == 'E')goal[i][j] = 0;
				else if (k == 'W')goal[i][j] = 3;
				else if (k == 'R')goal[i][j] = 1;
				else if (k == 'B')goal[i][j] = 2;
				if (ex == i && ey == j) {
					start[i][j].front = 0; start[i][j].top = 0;
				}
				else {
					start[i][j].front = 1; start[i][j].top = 3;
				}
			}
		dfs(-1, -1, 0);
		step = step > 30 ? -1 : step;
		cout << step << endl;
	}
	return 0;
}