POJ 3126 Prime Path【从一个素数变为另一个素数的最少步数/BFS】

Prime Path

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 26475 Accepted: 14555

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

【题意】：给你两个素数，s和e，要求在最少次数从s转变到e，而且中间的数字也得是素数，并且变化前后相邻的两个数只有一位不同

【分析】：是一道在素数集合上的搜索题目，并且一定是四位数的素数。题目要求所经过的路径最短，自然是BFS。关键是BFS构造的问题，有一个很简单的方法，就是把BFS想成一棵树，每个节点表示一个状态，这个节点的子节点则表示由该状态可到达的状态，对于一个四位数，比如1033，它的千位可以从1-9取值，百位和十位可以从0-9取值，而个位只能取奇数位，因为个位为偶数的数肯定不是素数，然后就是让所有可以到达的状态进队列，并且给他们标记，防止下次再次进入，至于判断素数，可以在O(1)内完成，所以整个程序的效率是比较高的。

最后如果弹出的数就是想要到达的数，输出步数

Tips：一般这种要求步数的BFS都是需要使用结构体，每一个节点记录自己的步数

【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const int N = 1e4+10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[][3]={ {0,0,1},{0,0,-1},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0} };
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

struct node
{
    int p[4],step; //數組代表4個數位的取值
}st,ed;

int vis[15000];
int t,n,s,e;
int a,b,c,d;

bool prime( int x )
{
    for(int i=2; i<=sqrt(x); i++)
        if(x%i==0)
            return false;
    return true;
}

int cal(int *a)
{
    return a[0]*1000 + a[1]*100 + a[2]*10 + a[3];
}

void bfs()
{
    memset(vis,false,sizeof(vis));
    queue<node>q;
    while(!q.empty()) q.pop();

    st.p[0]=s/1000; st.p[1]=s/100%10; st.p[2]=s/10%10; st.p[3]=s%10;
    st.step=0;
    q.push(st);
    vis[cal(st.p)] = 1;

    while(!q.empty())
    {
        st = q.front();
        q.pop();
        if(cal(st.p)==e)
        {
            printf("%d\n",st.step);
            return ;
        }
        for(int i=0;i<4;i++)//枚舉數位
        {

            for(int j=0; j<10; j++) //枚舉數位上值
            {
                ed=st;//
                if(i==0 && j==0) continue; //首位不能为0
                ed.p[i]=j; //给某位赋值
                if(!vis[cal(ed.p)] && prime(cal(ed.p)))
                {
                    vis[cal(ed.p)] = 1;
                    ed.step = st.step + 1;
                    q.push(ed);
                }
            }

        }
    }
   printf("Impossible\n");
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&s,&e);
        bfs();
    }
}
/*
3
1033 8179
1373 8017
1033 1033

6
7
0
*/