kuangbin 简单搜索 F题

F - 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

题解：

一共不到10000个数，暴力枚举每个数就好，是素数的话推进队列里。
TLE了一次啊。加深了对bfs的理解。
四状态Bfs+筛法素数表。
对每个已访问过的状态我们不应该再次访问，加个vis数组就过了。
注意千位不能出现0的情况。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;struct Node{   int data,times;   Node(int d,int ts):data(d),times(ts) {}};const int MAXN = 10000;bool vis[MAXN];bool notprime[MAXN];void init(){    memset(notprime,false,sizeof(notprime));    notprime[0]=notprime[1]=true;    for(int i=2;i<MAXN;i++)    {        if(!notprime[i])        {            if(i>MAXN/i)                continue;        }        for(int j=i*i;j<MAXN;j+=i)        {            notprime[j]=true;        }    }}int a,b;void bfs(){    queue<Node> que;    que.push(Node(a,0));    while(que.size())    {        Node p = que.front();        que.pop();        if(p.data==b)        {            //printf("%d\n",p.data);            printf("%d\n",p.times);            return;        }        int gewei = p.data%10;        int shiwei = p.data/10%10;        int baiwei = p.data/100%10;        int qianwei = p.data/1000;        //个位的变换        for(int i=1;i<10;i++)        {            int tmp = qianwei*1000+baiwei*100+shiwei*10+(gewei+i)%10;            if(!notprime[tmp]&&!vis[tmp])            {                vis[tmp]=1;                que.push(Node(tmp,p.times+1));            }        }        //十位的变换        for(int i=1;i<10;i++)        {            int tmp = qianwei*1000+baiwei*100+(shiwei+i)%10*10+gewei;            if(!notprime[tmp]&&!vis[tmp])            {                vis[tmp]=1;                que.push(Node(tmp,p.times+1));            }        }        //百位的变换        for(int i=1;i<10;i++)        {           int tmp = qianwei*1000+(baiwei+i)%10*100+shiwei*10+gewei;             if(!notprime[tmp]&&!vis[tmp])            {                vis[tmp]=1;                que.push(Node(tmp,p.times+1));            }        }        //千位的变换        for(int i=1;i<10;i++)        {            if(qianwei+i==10) continue;            int tmp = (qianwei+i)%10*1000+baiwei*100+shiwei*10+gewei;             if(!notprime[tmp]&&!vis[tmp])            {                vis[tmp]=1;                que.push(Node(tmp,p.times+1));            }        }    }     printf("Impossible.\n");}int main(){    int T;    scanf("%d",&T);    init();    while(T--)    {       memset(vis,false,sizeof(vis));       scanf("%d%d",&a,&b);       bfs();    }    return 0;}