Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 987 Accepted Submission(s): 635
Problem 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.
1033
1733
3733
3739
3779
8779
8179
The 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.
— 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.
1033
1733
3733
3739
3779
8779
8179
The 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.
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
Source
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wangye
题意:给你两个四位数,都是素数,每次改变可以改变其中的任何一个数字,但要求改变后的四位数(没有前导零)依然是素数,问最少改变几次可以使得第一个数改为第二个数
思路:可以先用埃氏筛法把1000-9999的所有的素数都选出来,之后就四个位数,每个位数最多改变八次,就搜索一下,我开了isprime和step两个数组,当然也可以开一个结构体,但只改变一个数字需要花点功夫,我是枚举了个位,十位,百位,千位。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<cstdlib> #include<queue> #include<set> #include<vector> using namespace std; #define INF 0x3f3f3f3f #define eps 1e-10 #define PI acos(-1.0) #define ll long long int const maxn = 9999; const int mod = 1e9 + 7; int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } bool isprime[maxn]; int step[maxn]; bool vis[maxn]; int num1,num2; void getprime(int n) { for(int i=0;i<=n;i++) isprime[i]=true; isprime[0]=isprime[1]=false; for(int i=2;i<=n;i++) { for(int j=2*i;j<=n;j=j+i) isprime[j]=false; } } int bfs(int st ) { vis[st]=true; step[st]=0; queue<int>que; que.push(st); while(que.size()) { int p=que.front(); que.pop(); if(p == num2) { return step[num2]; break; } int ge=p % 10; int shi=(p/10)%10; int bai=(p/100)%10; int qian=p/1000; for(int i=0;i<=9;i++) { int next; if(i!=ge) { next=p-ge+i; if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false) { step[next]=step[p]+1; que.push(next); vis[next]=true; // cout<<"ge"; } } if(i!=shi) { next=p-shi*10+i*10; if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false) { step[next]=step[p]+1; que.push(next); vis[next]=true; // cout<<"shi"; } } if(i!=bai) { next=p-bai*100+i*100; if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false) { step[next]=step[p]+1; que.push(next); vis[next]=true; // cout<<"bai"; } } if(i!=qian) { next=p-qian*1000+i*1000; if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false) { step[next]=step[p]+1; que.push(next); vis[next]=true; // cout<<"qian"; } } } } return -1; } int main() { int t; scanf("%d",&t); getprime(9999); while(t--) { memset(vis,false,sizeof(vis)); memset(step,0,sizeof(step)); scanf("%d %d",&num1,&num2); // for(int i=2;i<=9999;i++) // if(isprime[i]) // cout<<i<<" "; int ans=bfs(num1); if(ans>=0) printf("%d ",ans); else printf("Impossible "); } }