先上题目
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9259 | Accepted: 5274 |
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
One
line with a positive number: the number of test cases (at most 100).
Then for each test case, one line with two numbers separated by a blank.
Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
题意比较简单，就是给你两个4位的素数，问你能不能经过有限的步骤将第一个素数转化为第二个素数，其中转化的要符合一下的要求：每一次只可以改变这个四位数的某一位，首位不能为0，每一次转化以后得到的数要还是素数。如果不可以转化得到目标素数，输出Impossible。
做法也是比较简单，先筛出10000以内的素数出来，然后枚举当前的这个素数可以转化的其他素数，然后进行bfs，当遇到目标素数的时候就输出步数；如果转化不了就会把可以转化的素数都转化了一遍，最后循环就会结束。这里为了能够判断是否能转化，需要标记它转化过哪些素数，转化过的素数如果再出现，就不需要再搜索这个数了。
上代码：
1 #include <stdio.h> 2 #include <string.h> 3 #include <queue> 4 #include <map> 5 #define LL long long 6 #define MAX (10000+10) 7 using namespace std; 8 9 bool f[MAX],M[MAX]; 10 11 typedef struct 12 { 13 int l; 14 int st; 15 }S; 16 queue<S> q; 17 18 void dedeal() 19 { 20 LL i,j,n; 21 n=MAX; 22 for(i=2;i<=n;i++) 23 { 24 if(!f[i]) 25 { 26 for(j=i*i;j<=n;j+=i) f[j]=1; 27 } 28 } 29 } 30 31 int check(int x,int y) 32 { 33 int t,i; 34 S d; 35 d.l=x; 36 d.st=0; 37 memset(M,0,sizeof(M)); 38 while(!q.empty()) q.pop(); 39 q.push(d); 40 M[d.l]=1; 41 while(!q.empty()) 42 { 43 d=q.front(); 44 q.pop(); 45 if(d.l==y) return d.st; 46 d.st++; 47 x=d.l; 48 t=x/10*10; 49 for(i=0;i<10;i++) 50 { 51 d.l=t+i; 52 if(!f[d.l] && d.l!=x && !M[d.l]) 53 { 54 M[d.l]=1; 55 q.push(d); 56 } 57 } 58 t=x/100*100+x%10; 59 for(i=0;i<100;i+=10) 60 { 61 d.l=t+i; 62 if(!f[d.l] && d.l!=x && !M[d.l]) 63 { 64 M[d.l]=1; 65 q.push(d); 66 } 67 } 68 t=x/1000*1000+x%100; 69 for(i=0;i<1000;i+=100) 70 { 71 d.l=t+i; 72 if(!f[d.l] && d.l!=x && !M[d.l]) 73 { 74 M[d.l]=1; 75 q.push(d); 76 } 77 } 78 t=x%1000; 79 for(i=1000;i<10000;i+=1000) 80 { 81 d.l=t+i; 82 if(!f[d.l] && d.l!=x && !M[d.l]) 83 { 84 M[d.l]=1; 85 q.push(d); 86 } 87 } 88 } 89 return -1; 90 } 91 92 int main() 93 { 94 int n,c,x,y; 95 //freopen("data.txt","r",stdin); 96 dedeal(); 97 scanf("%d",&n); 98 while(n--) 99 { 100 scanf("%d %d",&x,&y); 101 c=check(x,y); 102 if(c==-1) printf("Impossible "); 103 else printf("%d ",c); 104 } 105 return 0; 106 }