Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14539 | Accepted: 8196 |
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： 输入多组数据，每组数据包括两个四位素数1033 8179，每次只改变四位数中的一位并且改变后的数也为素数，从1033到8179有6部。没有路径则输出Imbossiblei。
做法： 这是一个40端口的bfs不过剪枝之后就没有40入口了，入口数远小于40
无论是判定素数还是搜索素数，首先排除偶数，这样就剪掉一半枝叶了
判断素数用根号法判断，
如果一个数X不能被 [2,√X] 内的所有素数整除，那么它就是素数
可以判断的复杂度降到logn
注意：千位的变换要保证千位不为0
其实素数也是用来辅助搜索剪枝的#include <iostream>#include <stdio.h>
#include <queue> #include <string.h> using namespace std; const int MAX=10000; int dis[MAX],str[4]; bool vis[MAX]; bool just(int n) { for(int i=2; i*i<=n; i++) { if(n%i==0) return 0; } return 1; } int bfs(int star,int ends) { int y; memset(vis,0,sizeof(vis)); memset(dis,0,sizeof(dis)); queue<int> q; q.push(star); vis[star]=1; if(star==ends) return 0; while(!q.empty()) { int x=q.front(); q.pop(); str[0]=x/1000; str[1]=x/100%10; str[2]=x/10%10; str[3]=x%10; for(int i=0; i<4; i++) { int h=str[i];//注意这里记住这个str[i]； if(i==0) for(int j=1; j<10; j++) { str[i]=j; y=str[0]*1000+str[1]*100+str[2]*10+str[3]; if(!vis[y]&&just(y)) { q.push(y); vis[y]=1; dis[y]=dis[x]+1; if(y==ends) return dis[y]; } } else for(int j=0; j<10; j++) { str[i]=j; y=str[0]*1000+str[1]*100+str[2]*10+str[3]; if(!vis[y]&&just(y)) { q.push(y); vis[y]=1; dis[y]=dis[x]+1; if(y==ends) return dis[y]; } } str[i]=h;//在这里返回去；
} } return -1; } int main() { int t,star,ends; scanf("%d",&t); getchar(); while(t--) { scanf("%d%d",&star,&ends); int s=bfs(star,ends); if(s!=-1) printf("%d ",s); else printf("Impossible "); } return 0; }