挺不错的题目，很锻炼代码能力和调试能力~

题意：初始格子状态固定，给你移动后格子的状态，问最少需要多少步能到达，如果步数大于30，输出-1。

由于单向搜索状态太多，搜到二十几就会爆了，所以应该想到双向广搜。

对于每一个格子，我只需要记录上面和前面的格子颜色，因为格子左右移动前面颜色不变上面颜色变了，前后移动的话只是前面和上面的颜色交换了而已，不过写的太挫了。。。应该直接进制压缩一下。。。我还用了两个数来表示前面和上面的颜色，下次把这些搜索题全A了后再回来重新写一下，重写效果也会不错，并不是题目a了这题就好了。

#include <stdio.h>
#include <string.h>#define LL __int64const int mod = 100007;
struct HASH{int head[mod+10], E, next[222222];int v1[222222], v2[222222], c[222222];void init() {memset(head, -1, sizeof(head));E = 0;}int gethash(int x,int y) {return (((LL)x + y)%mod+mod)%mod;}int isin(int x, int y) {int u = gethash(x, y);for(int i = head[u];i != -1;i = next[i]) {if(v1[i] == x && v2[i] == y) return c[i];}return -1;}void push(int x, int y, int step) {int u = gethash(x, y);v1[E] = x;v2[E] = y;c[E] = step;next[E] = head[u];head[u] = E++;}
}ha1, ha2;struct PP{int st1, st2;int step;PP(){}PP(int st1, int st2, int step) : st1(st1), st2(st2), step(step) {}
}cur, to, q[1000002];int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int a[22], b[22], d[2222];void swap(int &a, int &b) {int t = a; a = b; b = t;
}void get_zt(int st1, int st2) {for(int i = 9;i >= 1;i --)  a[i] = st1%10, st1 /= 10;for(int i = 9;i >= 1;i --)  b[i] = st2%10, st2 /= 10;
}void get_xy(int &st1, int &st2) {st1 = st2 = 0;for(int i = 1;i <= 9; i++)  st1 = st1*10+a[i];for(int i = 1;i <= 9; i++)  st2 = st2*10+b[i];
}bool judge(int go) {int ret = 0;for(int i = 1;i <= 9; i++)  ret = ret*10+a[i];return go == ret;
}int getnew(int a, int b) {int vis[5], i;for(i = 1;i <= 3; i++) vis[i] = 0;vis[a]++;vis[b]++;for(i = 1;i <= 3; i++)  if(!vis[i]) return i;
}int min(int a, int b) {return a>b?b:a;
}int bfs(int st1,int st2, int go) {ha1.init();ha1.push(st1, st2, 0);int head = 0, tail = 0;q[tail++] = PP(st1, st2, 0);int i, j, k;while(head < tail) {cur = q[head++];st1 = cur.st1, st2 = cur.st2;get_zt(st1, st2);int now ;for(i = 1;i <= 9; i++)  if(a[i] == 4) {now = i; break;}int x = now/3+1, y = now%3;if(y == 0)  y = 3, x--;for(i = 0;i < 4; i++) {int xx = x+dx[i];int yy = y+dy[i];if(xx < 1 || xx > 3 || yy < 1 || yy > 3)    continue;int then = (xx-1)*3+yy;swap(a[now], a[then]);swap(b[now], b[then]);if(i < 2) {swap(a[now], b[now]);if(judge(go)) {return cur.step+1;}get_xy(to.st1, to.st2);to.step = cur.step+1;if(ha1.isin(to.st1, to.st2) == -1 && to.step <= 20) {ha1.push(to.st1, to.st2, to.step);if(to.step > 20)   continue;q[tail++] = to;}swap(a[now], b[now]);}else {a[now] = getnew(a[now], b[now]);if(judge(go)) {return cur.step+1;}get_xy(to.st1, to.st2);to.step = cur.step+1;if(ha1.isin(to.st1, to.st2) == -1 && to.step <= 20) {ha1.push(to.st1, to.st2, to.step);if(to.step > 20)   continue;q[tail++] = to;}a[now] = getnew(a[now], b[now]);}swap(a[now], a[then]);swap(b[now], b[then]);}}int temp , tot = 0;int N = 1<<8;for(i = 0;i < N; i++) {temp = go;for(j = 9;j >= 1; j--)a[j] = temp%10 , temp /= 10;int id1 = 1, id2 = 0, sum = 0;while(id1 <= 9) {if(a[id1] == 4) {sum = sum*10+4;id1++;continue;}if((1<<id2) & i) {for(k = 3;k >= 1; k--)   if(k != a[id1])    break;sum = sum*10+k;}else {for(k = 1;k <= 3; k++)   if(k != a[id1])    break;sum = sum*10+k;}id1++, id2++;}d[tot++] = sum;}for(i = 0;i < tot; i++){int now = ha1.isin(go, d[i]);if(now >= 0)    return now;}int ans = 1111111;for(int ii = 0;ii < tot;ii++) {ha2.init();cur = PP(go, d[ii], 0);int head = 0, tail = 0;q[tail++] = cur;int ok = 0;while(head < tail) {cur = q[head++];st1 = cur.st1, st2 = cur.st2;get_zt(st1, st2);int now ;for(i = 1;i <= 9; i++)  if(a[i] == 4) {now = i; break;}int x = now/3+1, y = now%3;if(y == 0)  y = 3, x--;for(i = 0;i < 4; i++) {int xx = x+dx[i];int yy = y+dy[i];if(xx < 1 || xx > 3 || yy < 1 || yy > 3)    continue;int then = (xx-1)*3+yy;swap(a[now], a[then]);swap(b[now], b[then]);if(i < 2) {swap(a[now], b[now]);get_xy(to.st1, to.st2);to.step = cur.step+1;int temp = ha1.isin(to.st1, to.st2);if(temp >= 0)  {ok = 1;ans = min(ans, temp+to.step);continue;}if(ha2.isin(to.st1, to.st2) == -1 && to.step <= 10) {ha2.push(to.st1, to.st2, to.step);q[tail++] = to;}swap(a[now], b[now]);}else {a[now] = getnew(a[now], b[now]);get_xy(to.st1, to.st2);to.step = cur.step+1;int temp = ha1.isin(to.st1, to.st2);if(temp >= 0)  {ok = 1;ans = min(ans, temp+to.step);continue;}if(ha2.isin(to.st1, to.st2) == -1 && to.step <= 10) {ha2.push(to.st1, to.st2, to.step);q[tail++] = to;}a[now] = getnew(a[now], b[now]);}swap(a[now], a[then]);swap(b[now], b[then]);if(ok)  break;}if(ok)  break;}}if(ans > 30)  return -1;return ans;
}int main() {int i, x, n, m;while(scanf("%d%d", &m, &n) != -1 && n) {int go = 0, st1 = 0, st2 = 0;char ch[2];for(i = 1;i <= 9; i++) {scanf("%s", ch);if(ch[0] == 'W') x = 1;else if(ch[0] == 'R')    x = 2;else if(ch[0] == 'B')    x = 3;else    x = 4;go = go*10 + x;if(i == (n-1)*3 + m) {st1 = st1*10+4;st2 = st2*10+4;}else  {st1 = st1*10+1;st2 = st2*10+2;}}if(st1 == go) {puts("0"); continue;}printf("%d\n", bfs(st1, st2, go));}return 0;
}