| Time Limit: 15000MS | Memory Limit: 150000KB | 64bit IO Format: %I64d & %I64u |
Description
The rotation game uses a # shaped board, which can hold 24 pieces of square blocks (see Fig.1). The blocks are marked with symbols 1, 2 and 3, with exactly 8 pieces of each kind.
Initially, the blocks are placed on the board randomly. Your task is to move the blocks so that the eight blocks placed in the center square have the same symbol marked. There is only one type of valid move, which is to rotate one of the four lines, each consisting of seven blocks. That is, six blocks in the line are moved towards the head by one block and the head block is moved to the end of the line. The eight possible moves are marked with capital letters A to H. Figure 1 illustrates two consecutive moves, move A and move C from some initial configuration.
Initially, the blocks are placed on the board randomly. Your task is to move the blocks so that the eight blocks placed in the center square have the same symbol marked. There is only one type of valid move, which is to rotate one of the four lines, each consisting of seven blocks. That is, six blocks in the line are moved towards the head by one block and the head block is moved to the end of the line. The eight possible moves are marked with capital letters A to H. Figure 1 illustrates two consecutive moves, move A and move C from some initial configuration.
Input
The input consists of no more than 30 test cases. Each test case has
only one line that contains 24 numbers, which are the symbols of the
blocks in the initial configuration. The rows of blocks are listed from
top to bottom. For each row the blocks are listed from left to right.
The numbers are separated by spaces. For example, the first test case in
the sample input corresponds to the initial configuration in Fig.1.
There are no blank lines between cases. There is a line containing a
single `0' after the last test case that ends the input.
Output
For each test case, you must output two lines. The first line contains
all the moves needed to reach the final configuration. Each move is a
letter, ranging from `A' to `H', and there should not be any spaces
between the letters in the line. If no moves are needed, output `No
moves needed' instead. In the second line, you must output the symbol of
the blocks in the center square after these moves. If there are several
possible solutions, you must output the one that uses the least number
of moves. If there is still more than one possible solution, you must
output the solution that is smallest in dictionary order for the letters
of the moves. There is no need to output blank lines between cases.
Sample Input
1 1 1 1 3 2 3 2 3 1 3 2 2 3 1 2 2 2 3 1 2 1 3 3 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 0
Sample Output
AC 2 DDHH 2
Source
略可怕的题，肝了两个多小时，最后还是看了提示才过。
这种翻来倒去求操作方式的题明显是搜索题，因为这道题里状态实在不好保存，没法BFS，那么就选DFS。我开了个二维数组，像存图一样模拟井字存了数字，光读入就用了好多行
你以为到这里就结束了吗？不！
这个井字总共有8种操作方式，且没有统一的公式，得把8种都写上！为了状态回溯，还得写8行求回溯操作的代码。另外判断是否符合条件还要写8行。这些处理完以后，就可以开始搜索了。
你以为之后就简单了吗？不！才刚刚开始！
开始搜索吧，先写好状态改变、存解、判断是否出解、进入深层搜索、状态回溯……很好，程序陷入死循环了，DFS撑不住这题超大的深度。
那么剪枝吧，在尝试了类似限制连续同种操作次数、限制总深度、用随机数限制深度等方法之后，最终想到了迭代加深搜索。啊，迭代加深太棒了，很快就出了解。
你以为可以AC了吗？不！麻烦的还在后头！
虽然过了样例，但接着是花式TLE。这时我想起了隔壁Orion_Rigel的提示：A星算法。
TM我根本不会A星啊！果断百度搜A星，然而并没有找到好用的教程，仍旧一脸懵。最后怒看题解，噫，秒懂。总之就是根据“当前的操作状况和预估的需要操作次数”决策是否剪枝（看代码更好理解）
。于是把判断函数加上了预估功能（返回相等数字个数，而不是“是否都相等”）。
Accepted
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<cmath> 6 using namespace std; 7 int mp[8][8]; 8 int mini; 9 int ans[350]; 10 int ansnum=0; 11 int len=0; 12 int depth=0; 13 14 //dfs 15 int num_cnt[4];//数字出现次数 16 17 // 18 inline bool read(){//读入 19 scanf("%d",&mp[1][3]); 20 if(mp[1][3]==0)return false; 21 scanf("%d",&mp[1][5]); 22 scanf("%d%d",&mp[2][3],&mp[2][5]); 23 for(int i=1;i<=7;i++)scanf("%d",&mp[3][i]); 24 scanf("%d%d",&mp[4][3],&mp[4][5]); 25 for(int i=1;i<=7;i++)scanf("%d",&mp[5][i]); 26 scanf("%d%d",&mp[6][3],&mp[6][5]); 27 scanf("%d%d",&mp[7][3],&mp[7][5]); 28 return true; 29 } 30 void mv(int a){//模拟数字变换 31 switch(a){ 32 case 1:{//'A' 33 int temp=mp[1][3]; 34 for(int i=1;i<=6;i++)mp[i][3]=mp[i+1][3]; 35 mp[7][3]=temp; 36 break; 37 } 38 case 2:{//'B' 39 int temp=mp[1][5]; 40 for(int i=1;i<=6;i++)mp[i][5]=mp[i+1][5]; 41 mp[7][5]=temp; 42 break; 43 } 44 case 3:{//'C' 45 int temp=mp[3][7]; 46 for(int i=7;i>=2;i--)mp[3][i]=mp[3][i-1]; 47 mp[3][1]=temp; 48 break; 49 } 50 case 4:{//'D' 51 int temp=mp[5][7]; 52 for(int i=7;i>=2;i--)mp[5][i]=mp[5][i-1]; 53 mp[5][1]=temp; 54 break; 55 } 56 case 5:{//'E' 57 int temp=mp[7][5]; 58 for(int i=7;i>=2;i--)mp[i][5]=mp[i-1][5]; 59 mp[1][5]=temp; 60 break; 61 } 62 case 6:{//'F' 63 int temp=mp[7][3]; 64 for(int i=7;i>=2;i--)mp[i][3]=mp[i-1][3]; 65 mp[1][3]=temp; 66 break; 67 } 68 case 7:{//'G' 69 int temp=mp[5][1]; 70 for(int i=1;i<=6;i++)mp[5][i]=mp[5][i+1]; 71 mp[5][7]=temp; 72 break; 73 } 74 case 8:{//'H' 75 int temp=mp[3][1]; 76 for(int i=1;i<=6;i++)mp[3][i]=mp[3][i+1]; 77 mp[3][7]=temp; 78 break; 79 } 80 } 81 return; 82 } 83 int pd(){//判断中间相同数字最多个数，8个即是相同了 84 memset(num_cnt,0,sizeof(num_cnt)); 85 num_cnt[mp[3][3]]++; 86 num_cnt[mp[3][4]]++; 87 num_cnt[mp[3][5]]++; 88 num_cnt[mp[4][3]]++; 89 num_cnt[mp[4][5]]++; 90 num_cnt[mp[5][3]]++; 91 num_cnt[mp[5][4]]++; 92 num_cnt[mp[5][5]]++; 93 return max(num_cnt[1],max(num_cnt[2],num_cnt[3]));//返回中心出现最多的数 94 } 95 int reset(int i){//返回还原操作的序号 96 if(i<=4){ 97 if(i==1)return 6; 98 if(i==2)return 5; 99 if(i==3)return 8; 100 if(i==4)return 7; 101 } 102 else{ 103 if(i==5)return 2; 104 if(i==6)return 1; 105 if(i==7)return 4; 106 if(i==8)return 3; 107 } 108 return 0; 109 } 110 bool dfs(int t,int last){ 111 if(depth-t<8-pd()) return false; 112 //预测值剪枝 113 if(t>=depth) return false; 114 //最优值剪枝 115 for(int i=1;i<=8;i++){ 116 if(i==reset(last))continue;//防止和上一步反向移动 117 mv(i);//移动 118 ans[t]=i; 119 if(pd()==8){//出解 120 len=t; 121 ansnum=mp[3][3]; 122 return true; 123 } 124 if(dfs(t+1,i))return true;//下一层 125 mv(reset(i));//还原 126 } 127 return false; 128 } 129 int main(){ 130 while(read()){ 131 int i,j; 132 if(pd()==8){ 133 printf("No moves needed "); 134 printf("%d ",mp[3][3]); 135 } 136 else{ 137 depth=1; 138 while(1){ 139 if(dfs(1,-1))break; 140 // printf("Dep:%d ",depth); 141 depth++;//迭代加深 142 } 143 for(int i=1;i<=len;i++)printf("%c",(char)ans[i]+'A'-1); 144 printf(" "); 145 printf("%d ",ansnum); 146 } 147 } 148 return 0; 149 }