F - Prime Path
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.
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 1033Sample Output
6
7
0AC代码1：
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10050;
const int INF = 0x3f3f3f3f;
const int mod = 7;
typedef long long ll;
#define PI 3.1415927
int n, m;
int t;
bool isPrime[maxn];
bool vis[maxn];
bool flag = false;
int ans = INF;
struct Node
{
int num[4];
int step;
};
void bfs(int x)
{
Node star;
int cnt = 3;
while(x)
{
star.num[cnt--] = x%10;
x/=10;
}
star.step = 0;
queue<Node> que;
que.push(star);
while(!que.empty())
{
Node next = que.front();
que.pop();
cnt = 3;
int nu = next.num[0]*1000+next.num[1]*100+next.num[2]*10+next.num[3];
if(nu == m)
{
ans = next.step;
break;
}
for(int i = 0; i < 4; ++i)
{
for(int j = 0; j <=9; j++)
{
if((i == 0 && j == 0) || (next.num[i]==j))
{
continue;
}
else
{
int numb = 0;
int temp = 1000;
for(int k = 0; k < 4; ++k)
{
if(k!=i)
{
numb += next.num[k]*temp;
temp /= 10;
}
else
{
numb += j*temp;
temp /= 10;
}
}
if(isPrime[numb] && !vis[numb])
{
vis[numb] = true;
cnt = 3;
while(numb)
{
star.num[cnt--] = numb%10;
numb/=10;
}
star.step = next.step+1;
que.push(star);
}
}
}
}
}
}
int main()
{
memset(isPrime, true, sizeof(isPrime));
for(int i = 2; i < maxn; ++i)
{
if(isPrime[i])
{
for(int j = i*i; j < maxn; j+=i)
{
isPrime[j] = false;
}
}
}
scanf("%d", &t);
while(t--)
{
ans = 0;
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
if(n == m)
{
printf("0\n");
continue;
}
bfs(n);
printf("%d\n", ans);
}
return 0;
}
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AC代码2：
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10050;
const int INF = 0x3f3f3f3f;
const int mod = 7;
typedef long long ll;
#define PI 3.1415927
int n, m;
int t;
bool isPrime[maxn];
bool vis[maxn];
bool flag = false;
int ans = INF;
struct Node
{
int num[4];
int step;
};
void bfs(int x)
{
Node star;
int cnt = 3;
while(x)
{
star.num[cnt--] = x%10;
x/=10;
}
star.step = 0;
queue<Node> que;
que.push(star);
while(!que.empty())
{
Node next = que.front();
que.pop();
cnt = 3;
int nu = next.num[0]*1000+next.num[1]*100+next.num[2]*10+next.num[3];
if(nu == m)
{
ans = next.step;
break;
}
for(int i = 0; i < 4; ++i)
{
for(int j = 0; j <=9; j++)
{
if((i == 0 && j == 0) || (next.num[i]==j))
{
continue;
}
else
{
int numb = 0;
int temp = 1000;
for(int k = 0; k < 4; ++k)
{
if(k!=i)
{
numb += next.num[k]*temp;
temp /= 10;
}
else
{
numb += j*temp;
temp /= 10;
}
}
if(isPrime[numb] && !vis[numb])
{
vis[numb] = true;
cnt = 3;
while(numb)
{
star.num[cnt--] = numb%10;
numb/=10;
}
star.step = next.step+1;
que.push(star);
}
}
}
}
}
}
int main()
{
memset(isPrime, true, sizeof(isPrime));
for(int i = 2; i < maxn; ++i)
{
if(isPrime[i])
{
for(int j = i*i; j < maxn; j+=i)
{
isPrime[j] = false;
}
}
}
scanf("%d", &t);
while(t--)
{
ans = 0;
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
if(n == m)
{
printf("0\n");
continue;
}
bfs(n);
printf("%d\n", ans);
}
return 0;
}