文章标题 POJ 3126 ： Prime Path （BFS）

Prime Path

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：给你两个四位数a,b，然后将a变成b，每次只能改变a中的一个位数，且改变完的这个四位数还是素数，问最少的次数能将a变为b。不能变为b就输出Impossible
分析：用bfs.可以先将1000~9999中的素数求出来，标记一下，当前出队的数位tmp,将tmp的四个位数每次改变一个位数上的数，如果是素数，切还未被访问过，则放进队列中，直到出队的是所求的值，就将到达次数所用的次数返回。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int a,b;int prime[10005];int vis[100005];//标记已经访问过 int  cnt;int judge_prime(int a){//判断a是否为素数     for (int i=2;i*i<=a;i++){        if (a%i==0)return 0;    }    return 1;}void init (){//初始化，如果是素数的话就标记为1     cnt=0;    memset (prime,0,sizeof (prime));    memset (vis,0,sizeof (vis));    for (int i=1001;i<=9999;i+=2){        if (judge_prime(i)==0)continue;        prime[i]=1;    }}struct node {    int a;    int cost;};int bfs(int a,int b){    memset (vis,0,sizeof (vis));    if (a==b){        return 0;    }    queue <node> q;    node tmp;    vis[a]=1;//标记    tmp.a=a;    tmp.cost=0;    q.push(tmp);    while (!q.empty()){        node t = q.front();q.pop();        if (t.a==b) return t.cost;//找到就返回         node tt;        for (int i=0;i<=9;i++){//个位数的情况             tt.a=t.a/10*10+i;            tt.cost=t.cost +1;            if (prime[tt.a]&&!vis[tt.a]){//如果是素数，且还未被访问过就进入队列                 q.push(tt);                vis[tt.a]=1;            }        }        for (int i=0;i<=9;i++){//十位数情况             int s=t.a%10;            tt.a=t.a/100*100+i*10+s;            tt.cost=t.cost +1;            if (prime[tt.a]&&!vis[tt.a]){                q.push(tt);                vis[tt.a]=1;            }        }        for (int i=0;i<=9;i++){//百位数情况             int s=t.a%100;            tt.a=t.a/1000*1000+i*100+s;            tt.cost=t.cost +1;            if (prime[tt.a]&&!vis[tt.a]){                q.push(tt);                vis[tt.a]=1;            }        }        for (int i=1;i<=9;i++){//千位数情况，千位数不能为0，i从1开始             int s=t.a%1000;            tt.a=t.a/10000+i*1000+s;            tt.cost=t.cost +1;            if (prime[tt.a]&&!vis[tt.a]){                q.push(tt);                vis[tt.a]=1;            }        }    }    return -1;//没有返回-1 }int main (){    int t;    init();    cin>>t;    while (t--){        cin>>a>>b;        int ans = bfs(a,b);        if (ans==-1) cout<<"Impossible"<<endl;        else cout<<ans<<endl;    }     return 0;}