Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?20325	?	Accepted:?7830	?	Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location?[i, j]?(1 ≤ i, j ≤ 4). However, this also changes states of all handles in row?i?and all handles in column?j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “?” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+-- ---- ---- -+--

Sample Output

6 1 1 1 3 1 4 4 1 4 3 4 4 1 #include<stdio.h> 2 #include<string.h> 3 int main() { 4 int in[16]; 5 int tmp[16]; 6 int length=4; 7 int size=length*length; 8 char temp; 9 //將輸入變成0,1，開的為1，關的為0，放在數組in中 10 for(int i=0;i<size;i++){ 11 scanf("%c",&temp); 12 if(temp=='\n') 13 scanf("%c",&temp); 14 if(temp=='+') 15 in[i]=0; 16 else 17 in[i]=1; 18 } 19 // for(int i=0;i<size;i++){ 20 // printf("%d \n",in[i]); 21 // } 22 //com數組存放的是開關的組合情況 23 int com[16]; 24 for(int i=1;i<size+1;i++){ 25 26 for(int j=0;j<i;j++){ 27 com[j]=j; 28 } 29 int end=0; 30 while(1){ 31 memcpy(tmp,in,size*sizeof(int)); 32 //按照搜索到的組合情況對冰箱進行開關 33 for(int j=0;j<i;j++){ 34 int row=com[j]/4; 35 int col=com[j]%4; 36 for(int k=0;k<length;k++){ 37 tmp[row*length+k]=1-tmp[row*length+k]; 38 tmp[k*length+col]=1-tmp[k*length+col]; 39 } 40 tmp[row*length+col]=1-tmp[row*length+col]; 41 } 42 int zero=0; 43 for(int j=0;j<size;j++){ 44 if(tmp[j]==1){ 45 zero+=1; 46 } 47 } 48 // for(int k=0;k<4;k++){ 49 // printf("%d\t",com[k]); 50 // } 51 // printf("\n"); 52 // printf("sero is %d\n",zero); 53 // zero=size; 54 //判斷是否找到答案 55 if(zero==size){ 56 end=1; 57 break; 58 } 59 //判斷開關情況是i個的組合是否已經全部搜索完畢，是則搜索i+1個情況 60 if(com[0]==size-i) 61 break; 62 //搜索下一個開關為i個的組合情況 63 for(int j=i-1;j>=0;j--){ 64 if(com[j]!=size-(i-j)){ 65 com[j]++; 66 for(int k=j+1;k<i;k++) 67 com[k]=com[k-1]+1; 68 break; 69 } 70 71 } 72 73 } 74 //若找到結果，輸出結果 75 if(end==1){ 76 printf("%d\n",i); 77 for(int j=0;j<i;j++){ 78 printf("%d %d\n",com[j]/4+1,com[j]%4+1); 79 } 80 } 81 } 82 83 return 0; 84 }

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轉載于:https://www.cnblogs.com/sdxk/p/4587218.html

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