【ACM】【BFS】POJ 3126 Prime Path

Prime Path

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32181		Accepted: 17439

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

Source

Northwestern Europe 2006

题目大意：

内阁首相抱怨机密部门的机密工作做的太差、为了保证自己的安全，内阁首相要求他的机密部门的人、每天给自己换一个门牌号，而且必须是质数、只有四位数字且每次只能换一位，问：从质数a到质数b每次只能换一位数字、最快要多少次能从a换到b？

解题思路：

运用BFS搜索，把每种情况遍历一遍就可以了。就是暴力；把每种情况都遍历一遍、然后就得出答案了。每次遍历的时候一定要初始化！！！

解题时候遇到的问题：

解题过程中、我想尽可能的剪枝，所以我的next数组只存了几个质因子，但是第一次提交直接WA了、然后我就发现千位、百位、十位也可能是10个数字的任意一个，所以又把数字都给加上了，然后第二次提交就直接AC了，心情美美哒。

AC代码如下：（有点长）

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

typedef pair<int,int> point;
int minn=99999;
bool pp[9999]={false};
bool book[9999]={false};
int next[10]={0,1,2,3,4,5,6,7,8,9};
queue<point> que;
bool prime(int n)
{	int i;
	for(i=2;i<n/2+1;i++)
		if(n%i==0)
			return false;
	return true;
}
void isprime()
{
	for(int i=1001;i<=9999;i+=2)
		if(prime(i))
			pp[i]=true;
}

void qsw(int x,int &a,int &b,int &c,int &d)
{
	d=x%10;
	x/=10;
	c=x%10;
	x/=10;
	b=x%10;
	x/=10;
	a=x%10;
}

void solve()
{
	point p,p1;
	minn=99999;
	int Start,End;
	cin >> Start >> End;
	if(Start==End)
		cout << "0\n";
	else
	{
		for(int i=1000;i<9999;i++)
			book[i]=false;
		while(!que.empty())
			que.pop();
		bool flag=false;
		p.first=Start;
		p.second=0;
		book[Start]=true;
		que.push(p);
		while(!que.empty())
		{
			p=que.front();
			que.pop();
			int x=p.first;
			int g,s,b,q;
			qsw(x,q,b,s,g);
	//		cout << q << " " << b << " " << s << " " << g << endl;
			//遍历该数字的四位 
			//个位 
			for(int i=0;i<10;i++)
			{
				int t=q*1000+b*100+s*10+next[i];
				if(pp[t]&&book[t]==false)
				{
					book[t]=true;
					p1.first=t;
					p1.second=p.second+1;
					que.push(p1);
					if(t==End)
					{
						flag=true;
						break;
					}
				}
			}
			if(flag) break;
			//十位 
			for(int i=0;i<10;i++)
			{
				int t=q*1000+b*100+next[i]*10+g;
				if(pp[t]&&book[t]==false)
				{
					book[t]=true;
					p1.first=t;
					p1.second=p.second+1;
					que.push(p1);
					if(t==End)
					{
						flag=true;
						break;
					}
				}
				
			}
			if(flag) break;
			//百位
			for(int i=0;i<10;i++)
			{
				int t=q*1000+next[i]*100+s*10+g;
				if(pp[t]&&book[t]==false)
				{
					book[t]=true;
					p1.first=t;
					p1.second=p.second+1;
					que.push(p1);
					if(t==End)
					{
						flag=true;
						break;
					}	
				}	
			} 
			if(flag) break;
			//千位
			for(int i=0;i<10;i++)
			{
				int t=next[i]*1000+b*100+s*10+g;
				if(pp[t]&&book[t]==false)
				{
					book[t]=true;
					p1.first=t;
					p1.second=p.second+1;
					que.push(p1);
					if(t==End)
					{
						flag=true;
						break;
					}
				}
			 } 
			 if(flag) break;
		}
			
		if(flag==false)
			cout << "Impossible\n";
		else
		{
			p=que.back();
			cout << p.second << endl;
		}
	}
}

int main()
{
	isprime();
	int m;
	cin >> m;
	while(m--)
	{
		solve();
	}
	return 0;
}