[kuangbin带你飞]专题1 简单搜索 F - Prime Path POJ - 3126

题目：

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23902		Accepted: 13210

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

题意：给一个四位素数，问最早经过几次变化可以到达输入的另一个四位素数。若无法到达输出”Impossible“，否则输出步数。输入保证这两个四位素数没有前导0。

每次变化可以改变一个数位上的数字（隐含：千位上的数不能变成0）

思路：先用素数筛把9999以内的素数打表都找出来，从当前状态枚举所有可以一步到达的数字，若可变到的数字是素数就加入BFS队列。最先到达目标素数的步数就是最小步数。走过的数字要做标记避免TLE。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<stack>

typedef long long ll;

using namespace std;

bool Prime[10010];
bool use[10010];

void getPrime(){
	memset(Prime,1,sizeof(Prime));
	Prime[2] = 1;
	for(int i = 2 ; i*i<=9999 ; i++){
		if(Prime[i]){
			for(int j = i+i ; j<=9999 ; j+=i){
				Prime[j] = 0;
			}
		}
	}
}

int start,end;
int tempN;

struct node{
	int num;
	int step;
}; 

node t;
node nxt;

int main(){
	ios::sync_with_stdio(false);
	getPrime();

	int cases;
	cin>>cases;
	
	int ans;
	while (cases--){
		cin>>start>>end;
		if(start == end){
			cout<<0<<endl;
			continue;
		}
		memset(use,0,sizeof(use));
		use[start] = 1;
		
		t.num = start, t.step = 0;
		
		queue<node> q;
		q.push(t);
		ans = -1;
		while(!q.empty()){
			t = q.front(); q.pop();
			if(t.num == end){
				ans = t.step;
				break;
			}

			int qian = t.num/1000;
			int bai = t.num%1000 /100;
			int shi = t.num%100 /10;
			int ge = t.num%10;
			for(int i = 0 ; i<=9 ; i++){
				tempN = qian*1000 + i*100 + shi*10 + ge;
				if(Prime[tempN] && !use[tempN]){
					nxt.num = tempN; nxt.step = t.step+1;
					use[tempN] = 1;
					q.push(nxt);
				}
				tempN = qian*1000 + bai*100 + i*10 + ge;
				if(Prime[tempN] && !use[tempN]){
					nxt.num = tempN; nxt.step = t.step+1;
					use[tempN] = 1;
					q.push(nxt);
				}
				tempN = qian*1000 + bai*100 + shi*10 + i;
				if(Prime[tempN] && !use[tempN]){
					nxt.num = tempN; nxt.step = t.step+1;
					use[tempN] = 1;
					q.push(nxt);
				}
				if(i==0)	continue;   //千位上的数不能变成0 
				tempN = i*1000 + bai*100 + shi*10 + ge;
				if(Prime[tempN] && !use[tempN]){
					nxt.num = tempN; nxt.step = t.step+1;
					use[tempN] = 1;
					q.push(nxt);
				}
			}
		}
		if(ans == -1){
			cout<<"Impossible"<<endl;
		}
		else{
			cout<<ans<<endl;
		}
	} 
	return 0;
}