IOI 98
To brighten up the gala dinner of the IOI'98 we have a set of N (10 <= N <= 100) colored lamps numbered from 1 to N.
The lamps are connected to four buttons:
- Button 1: When this button is pressed, all the lamps change their state: those that are ON are turned OFF and those that are OFF are turned ON.
- Button 2: Changes the state of all the odd numbered lamps.
- Button 3: Changes the state of all the even numbered lamps.
- Button 4: Changes the state of the lamps whose number is of the form 3xK+1 (with K>=0), i.e., 1,4,7,...
A counter C records the total number of button presses.
When the party starts, all the lamps are ON and the counter C is set to zero.
You are given the value of counter C (0 <= C <= 10000) and the final state of some of the lamps after some operations have been executed. Write a program to determine all the possible final configurations of the N lamps that are consistent with the given information, without repetitions.
PROGRAM NAME: lamps
INPUT FORMAT
No lamp will be listed twice in the input.
| Line 1: | N |
| Line 2: | Final value of C |
| Line 3: | Some lamp numbers ON in the final configuration, separated by one space and terminated by the integer -1. |
| Line 4: | Some lamp numbers OFF in the final configuration, separated by one space and terminated by the integer -1. |
SAMPLE INPUT (file lamps.in)
10 1 -1 7 -1
In this case, there are 10 lamps and only one button has been pressed. Lamp 7 is OFF in the final configuration.
OUTPUT FORMAT
Lines with all the possible final configurations (without repetitions) of all the lamps. Each line has N characters, where the first character represents the state of lamp 1 and the last character represents the state of lamp N. A 0 (zero) stands for a lamp that is OFF, and a 1 (one) stands for a lamp that is ON. The lines must be ordered from least to largest (as binary numbers).
If there are no possible configurations, output a single line with the single word `IMPOSSIBLE'
SAMPLE OUTPUT (file lamps.out)
0000000000 0101010101 0110110110
In this case, there are three possible final configurations:
- All lamps are OFF
- Lamps 1, 4, 7, 10 are OFF and lamps 2, 3, 5, 6, 8, 9 are ON.
- Lamps 1, 3, 5, 7, 9 are OFF and lamps 2, 4, 6, 8, 10 are ON.
题解：每个操作最多执行一次，因为执行两次相当于没有执行。所以如果输入的操作总次数C大于4，那么反复使C减去2，直到C小于或等于4。我们只需对前六个灯进行操作，因为第六个灯之后的灯都是前六个灯的重复，因此在输入的最终状态时的时候可以mod 6。对于1-4的操作，我们可以使用位运算来实现。
按钮1：XOR （111111）2=63
按钮2：XOR （010101）2=42
按钮3：XOR （101010）2=21
按钮4：XOR （011011）2=36
进行DFS时每次搜的时候只有两种状态，要么进行当前操作，要么不执行。所以DFS很容易写出来。
View Code
1 /* 2 ID:spcjv51 3 PROG:lamps 4 LANG:C 5 */ 6 #include<stdio.h> 7 #define MAXSTEP 4 8 int c,n,ans,total; 9 int a[10],f[10],b[10]; 10 int compare(const void*a,const void*b) 11 { 12 return(*(int*)a-*(int*)b); 13 } 14 void chai(int tt) 15 { 16 int i; 17 i=0; 18 memset(b,0,sizeof(b)); 19 while(tt) 20 { 21 i++; 22 b[i]=tt&1; 23 tt>>=1; 24 } 25 } 26 void check(int sum,int t) 27 { 28 int i,tt; 29 memset(b,0,sizeof(b)); 30 if(t>c||(c-t)%2!=0) return; 31 chai(sum); 32 for(i=1; i<=6; i++) 33 { 34 if(a[6-i+1]==-1) continue; 35 if(a[6-i+1]!=b[i]) break; 36 } 37 if(i>6) 38 { 39 total++; 40 f[total]=sum; 41 } 42 } 43 void dfs(int step,int t) 44 { 45 if(step>MAXSTEP) 46 { 47 48 check(ans,t); 49 return; 50 } 51 t++; 52 if(step==1) ans=ans^63; 53 if(step==2) ans=ans^42; 54 if(step==3) ans=ans^21; 55 if(step==4) ans=ans^36; 56 dfs(step+1,t); 57 t--; 58 if(step==1) ans=ans^63; 59 if(step==2) ans=ans^42; 60 if(step==3) ans=ans^21; 61 if(step==4) ans=ans^36; 62 dfs(step+1,t); 63 } 64 void print() 65 { 66 int i,j; 67 for(i=1;i<=total;i++) 68 { 69 chai(f[i]); 70 int c[10]; 71 for(j=1;j<=6;j++) 72 c[6-j+1]=b[j]; 73 for(j=1; j<n; j++) 74 printf("%d",c[(j-1)%6+1]); 75 printf("%d\n",c[(j-1)%6+1]); 76 } 77 } 78 int main(void) 79 { 80 freopen("lamps.in","r",stdin); 81 freopen("lamps.out","w",stdout); 82 int i,m,j,temp; 83 scanf("%d",&n); 84 scanf("%d",&c); 85 memset(a,-1,sizeof(a)); 86 while(scanf("%d",&m)!=EOF&&m!=-1) 87 a[(m-1)%6+1]=1; 88 while(scanf("%d",&m)!=EOF&&m!=-1) 89 { 90 91 if(a[(m-1)%6+1]!=-1) 92 { 93 printf("IMPOSSIBLE\n"); 94 return 0; 95 } 96 a[(m-1)%6+1]=0; 97 } 98 while(c>4) c-=2; 99 ans=63; 100 total=0; 101 dfs(1,0); 102 qsort(f,total,sizeof(int),compare); 103 if(total==0) 104 printf("IMPOSSIBLE\n"); 105 else 106 print(); 107 return 0; 108 }