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 1033
Sample Output
6
7
0
题意：
给你两个四位数的素数，通过改变其中的一位数，每次只允许改变一位数，而且改变之后的数也必须是个素数，
问你最少通过改变几次变成后一个四位的素数。如果不能改变成后面的四位素数则输出Impossible。
关键：打印素数链
因为每次只能改变一个数字 所以要把只相差一个数字的4喂数的素数连接起来
bool judge(int n, int m){
int cnt = 0;
while(n){
if(n%10==m%10) cnt++;
n /= 10;
m /= 10;
}
if(cnt > 2) return true;
return false;
}
void getMap(){
for(int i = 1; i <= prime[0]; i++){
for(int j = i +1; j <= prime[0];j++){
if(judge(prime[i], prime[j])){
map[prime[j]].push_back(prime[i]);
map[prime[i]].push_back(prime[j]);
}
}
}
}
在打印素数链前先get到素数 就是把素数存起来
bool isPrime(int n){
for(int i = 2; i <= sqrt(n); i++){
if(n % i == 0)
return false;
}
return true;
}
void getPrime()
{
for(int i=1000;i<10000;i++)
{
if(isPrime(i))
prime[++prime[0]] = i;//这里的prime[0]表示素数的个数
}
}
完整代码
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
int is_prime[10010];
int prime[10010];
int t;
int n, m;
vector<int> map[10010];
int ans;
bool vis[10010];
struct Node{
int x;
int step;
};
bool isPrime(int n){
for(int i = 2; i <= sqrt(n); i++){
if(n % i == 0)
return false;
}
return true;
}
void getPrime()
{
for(int i=1000;i<10000;i++)
{
if(isPrime(i))
prime[++prime[0]] = i;
}
}
//如果只相差1个数字 就用邻接表添加
bool judge(int n, int m){
int cnt = 0;
while(n){
if(n%10==m%10) cnt++;
n /= 10;
m /= 10;
}
if(cnt > 2) return true;
return false;
}
void getMap(){
for(int i = 1; i <= prime[0]; i++){
for(int j = i +1; j <= prime[0];j++){
if(judge(prime[i], prime[j])){
map[prime[j]].push_back(prime[i]);
map[prime[i]].push_back(prime[j]);
}
}
}
}
int bfs(){
memset(vis, false, sizeof(vis));
queue<Node> q;
Node a, b;
a.step = 0;
a.x = n;
vis[a.x] = true;
q.push(a);
while(!q.empty()){
a = q.front();
q.pop();
if(a.x == m){
return a.step;
}
for(int i = 0; i < map[a.x].size(); i++){
int tem = map[a.x][i];
if(!vis[tem]){
vis[tem] = true;
b.x = tem;
b.step = a.step + 1;
q.push(b);
}
}
}
return -1;
}
int main(){
getPrime();
getMap();
scanf("%d", &t);
while(t--){
scanf("%d%d", &n,&m);
ans = bfs();
if(ans == -1)
cout << "Impossible" << endl;
else cout << ans << endl;
}
return 0;
}
