POJ 3126

#include<iostream>
#include<queue>
#include<string>
using namespace std;

const int M = 10000;
int prime[M],pn;
bool notprime[M];

void init()
{
	pn = 0;
	notprime[1] = 1;
	for(int i = 2; i < M; i++)
	{
		if(notprime[i] == 0)
			prime[pn++] = i;
		for(int j = 0; j < pn && prime[j]*i < M; j++)
		{
			notprime[prime[j]*i] = 1;
			if(i%prime[j] == 0) break;
		}	
	}
}

int vis[M];
struct node{
	int x,step;
};
queue<node>q;

int bfs(int n, int m)
{
	node first_node;
	first_node.step = 0;
	first_node.x = n;
	q.push(first_node);
	vis[n] = 1; //标记n走过 
	node next; 
	while(!q.empty())
	{
		node tmp = q.front();
		int x = tmp.x;
		q.pop();
		if(x == m) return tmp.step; //如果到了终点 
		int num_1 = x/1000, num_2 = (x%1000)/100, num_3 = (x-num_1*1000-num_2*100)/10, num_4 = x%10;
		//千位 
		for(int i = 1; i < 10; i++)
		{
			int num = i*1000 + num_2*100 + num_3*10 + num_4;
			if(!vis[num] && !notprime[num])
			{
				next.x = num;
				next.step = tmp.step + 1;
				q.push(next);
				vis[num] = 1;	
			}
		}
		//百位和十位 
		for(int i = 0; i < 10; i++)
		{
			int num = num_1*1000 + i*100 + num_3*10 + num_4;
			if(!vis[num] && !notprime[num])
			{
				next.x = num;
				next.step = tmp.step + 1;
				q.push(next);
				vis[num] = 1;	
			}
			
			 num = num_1*1000 + num_2*100 + i*10 + num_4;
			 if(!vis[num] && !notprime[num])
			{
				next.x = num;
				next.step = tmp.step + 1;
				q.push(next);
				vis[num] = 1;	
			}
		}
		//个位 
		for(int i = 1; i < 10; i = i+2)
		{
			int num = num_1*1000 + num_2*100 + num_3*10 + i;
			if(!vis[num] && !notprime[num])
			{
				next.x = num;
				next.step = tmp.step + 1;
				q.push(next);
				vis[num] = 1;	
			}
		}
	}
	return -1;
}





int main()
{
	init();
	int t;
	cin>>t;
	while(t--){
		while(!q.empty()) q.pop();
		for(int i = 0; i < M; i++) vis[i] = 0;
		int n,m;
		cin>>n>>m;
		int ans = bfs(n,m);
		if(ans == -1) 
			cout<<"Impossible"<<endl; 
		else 
			cout<<ans<<endl;
	}
	
	
	return 0;
}

 代码很长，思路不难。把四位数拿出来，循环看是不是素数。加上bfs就ok了。