题目
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.
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
代码
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
const int N=10005;
int p[N],ip[N],vis[N],s,e;
void gp()//线性筛求素数
{
for(int i=2;i<=N;i++){
if(!p[i]){
ip[i]=1;
for(int j=2;j*i<=N;j++){
p[i*j]=1;
}
}
}
}
bool check(int x)
{
return !vis[x]&&ip[x];
}
void bfs()
{
queue<int> q;
q.push(s);
int ans=0;
while(!q.empty())
{
int n=q.size();
for(int i=1;i<=n;i++){
int cur=q.front();
vis[cur]=1;
q.pop();
if(cur==e){
cout<<ans<<endl;
return;
}
for(int i=0;i<=9;i++){
//更换个位
int f=cur%10;
int next=cur-f+i;
if(check(next)){
q.push(next);
}
}
for(int i=0;i<=9;i++){
//更换十位
int f=(cur/10)%10;
int next=cur-f*10+i*10;
if(check(next)){
q.push(next);
}
}
for(int i=0;i<=9;i++){
//更换百位
int f=(cur/100)%10;
int next=cur-f*100+i*100;
if(check(next)){
q.push(next);
}
}
for(int i=1;i<=9;i++){
//更换千位，注意0不能做千位
int f=cur/1000;
int next=cur-f*1000+i*1000;
if(check(next)){
q.push(next);
}
}
}
ans++;
}
cout<<"Impossible"<<endl;
}
int main()
{
gp();
/*for(int i=1000;i<=10000;i++){
if(ip[i]) cout<<i<<endl;
}*/
int t;
cin>>t;
while(t--)
{
memset(vis,0,sizeof(vis));
cin>>s>>e;
bfs();
}
return 0;
}
/*
3
1033 8179
1373 8017
1033 1033
*/
总结：
很有意思的bfs题