poj 1166 简单搜索
The Clocks
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12508 | Accepted: 4949 |
Description
|-------| |-------| |-------|
| | | | | | |
|---O | |---O | | O |
| | | | | |
|-------| |-------| |-------|
A B C
|-------| |-------| |-------|
| | | | | |
| O | | O | | O |
| | | | | | | | |
|-------| |-------| |-------|
D E F
|-------| |-------| |-------|
| | | | | |
| O | | O---| | O |
| | | | | | | |
|-------| |-------| |-------|
G H I
(Figure 1)
There are nine clocks in a 3*3 array (figure 1). The goal is to return all the dials to 12 o'clock with as few moves as possible. There are nine different allowed ways to turn the dials on the clocks. Each such way is called a move. Select for each move a number 1 to 9. That number will turn the dials 90' (degrees) clockwise on those clocks which are affected according to figure 2 below.
Move Affected clocks
1 ABDE
2 ABC
3 BCEF
4 ADG
5 BDEFH
6 CFI
7 DEGH
8 GHI
9 EFHI
(Figure 2)
Input
Your program is to read from standard input. Nine numbers give the start positions of the dials. 0=12 o'clock, 1=3 o'clock, 2=6 o'clock, 3=9 o'clock.
Output
Your program is to write to standard output. Output a shortest sorted sequence of moves (numbers), which returns all the dials to 12 o'clock. You are convinced that the answer is unique.
Sample Input
3 3 0 2 2 2 2 1 2
Sample Output
4 5 8 9
思路：
非常简单的一道题啊。。。为什么我做了这么久咧。。。。
暴搜就可以过的，直接九层for循环，再加判断就行啦。本人还在想着怎么用BFS做，结果BFS出来还TLE。
看来以后的思路要拓宽些了。
因为无论是1~9哪个操作，只要知道进行了哪些操作，进行了几次，不需要管其中的顺序，就可以直接得到最后的结果，又因为每个操作
做四次就回到了原来，所以1~9的每个操作，最多做4次。这样一来，就是9重for循环，每重做4次。。。
代码实现：
View Code
1 #include<iostream> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 int a[2000000][10]; 6 int doing[9][9]={1,1,0,1,1,0,0,0,0, 1,1,1,0,0,0,0,0,0, 0,1,1,0,1,1,0,0,0, 1,0,0,1,0,0,1,0,0, 0,1,0,1,1,1,0,1,0, 7 0,0,1,0,0,1,0,0,1, 0,0,0,1,1,0,1,1,0, 0,0,0,0,0,0,1,1,1, 0,0,0,0,1,1,0,1,1}; 8 int main(){ 9 int now[10]; 10 int num[10]; 11 while(cin>>now[0]>>now[1]>>now[2]>>now[3]>>now[4]>>now[5]>>now[6]>>now[7]>>now[8]){ 12 memset(num,0,sizeof(num)); 13 int times=0; 14 for(int i1=0;i1<4;i1++){ 15 num[0]=i1; 16 for(int i2=0;i2<4;i2++){ 17 num[1]=i2; 18 for(int i3=0;i3<4;i3++){ 19 num[2]=i3; 20 for(int i4=0;i4<4;i4++){ 21 num[3]=i4; 22 for(int i5=0;i5<4;i5++){ 23 num[4]=i5; 24 for(int i6=0;i6<4;i6++){ 25 num[5]=i6; 26 for(int i7=0;i7<4;i7++){ 27 num[6]=i7; 28 for(int i8=0;i8<4;i8++){ 29 num[7]=i8; 30 for(int i9=0;i9<4;i9++){ 31 num[8]=i9; 32 int flag=0; 33 for(int i10=0;i10<9;i10++){ 34 int temp=now[i10]; 35 for(int j=0;j<9;j++){ 36 now[i10]+=(doing[j][i10]*num[j]); 37 38 } 39 40 if((now[i10]%4)!=0){ 41 flag=1; 42 } 43 now[i10]=temp; 44 } 45 if(flag==0){ 46 for(int k=0;k<9;k++){ 47 a[times][k]=num[k]; 48 } 49 50 times++; 51 } 52 } 53 } 54 } 55 } 56 } 57 } 58 } 59 } 60 } 61 int min=10000000; 62 int temp2=0; 63 for(int g=0;g<times;g++){ 64 int sum=0; 65 for(int i=0;i<9;i++){ 66 sum+=a[g][i]; 67 } 68 if(min>sum){ 69 min=sum; 70 temp2=g; 71 } 72 } 73 int time1=0; 74 for(int j=0;j<9;j++){ 75 int k; 76 for(k=0;k<a[temp2][j];k++){ 77 if(time1!=0) 78 cout<<" "; 79 cout<<j+1; 80 time1++; 81 } 82 } 83 cout<<endl; 84 } 85 return 0; 86 }
