POJ-3126 Prime Path(BFS 求最小步数)

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
分析
mdzz,这题犯了几个傻逼错误，写到半夜…每次转换应该是素数的，我开始也是这样想的，但中间因为 tmp=num[i]; num[i]=tmp;没有写，让我怀疑是不是每次转换不一定是素数，然后查了下素数表，神奇的是我当时查的时候没看见，以为不一定是素数，后来把刚才那个bug改了，能搜出来,但输出数字不对，然后又tm查素数表，居然查到了(:з」∠)。 我查了假表，做了假题…..

经修饰的代码

 /************************************************************************* > File Name: poj3126.cpp    > Author:LaPucelle    > Mail:     > Created Time: 2017年01月21日 星期六 23时41分38秒 ************************************************************************/#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstdlib>#include<climits>#include<string>#include<cstring>#include<vector>#include<set>#include<list>#include<map>#include<queue>#define rep(i,a,b) for(int i=(a);i<=(b);(i++))#define inf 0x3fff3f#define ll long long#define pi acos(-1)int dire[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };int dire2[8][2]={{-1,-1},{-1,0},{-1,1},{ 0,-1},{ 0,1},{ 1,-1},{ 1,0},{ 1,1}};int dire3[6][3]={ {0,0,1},{0,1,0},{1,0,0},{0,0,-1},{0,-1,0},{-1,0,0} };using namespace std;const int maxn=10000;int prime[maxn];int vis[maxn];//判断是否为素数inline int isprime(int n){    int now=0;    for(int i=2;i*i<=n;i++){        if(n%i==0){            now=1;            break;        }    }    if(now)        return 0;    return 1;}// 打表int gene(int l,int r){    for(int i=l;i<=r;i++){        if(isprime(i)){            prime[i]=1;        }    }}int step[maxn];//记录走的步数// bfs void bfs(int n,int m){    queue<int> q;    q.push(n);    step[n]=0;    vis[n]=1;    while(!q.empty() ){        int head=q.front();        q.pop();        if(head==m){            printf("%d\n",step[m]);            return;        }        int num[5];        num[1]=head/1000;        num[2]=head/100%10;        num[3]=head/10%10;        num[4]=head%10;        int next;        for(int i=4;i>=1;i--){            int j;            if(i==1)                j=1;            else                j=0;            int tmp=num[i];            for(;j<=9;j++){                num[i]=j;                next=num[1]*1000+num[2]*100+num[3]*10+num[4];                if( prime[next] && !vis[next]){                    vis[next]=1;                    q.push(next);                    step[next]=step[head]+1;                }            }            num[i]=tmp; //容易错的地方        }            }}int main(){    //freopen("in.txt","r",stdin);    memset(prime,0,sizeof(prime));    gene(1000,9999);    int t;    cin>>t;    while(t--){        int n,m;        cin>>n>>m;        memset(vis,0,sizeof(vis));        memset(step,0,sizeof(step));        bfs(n,m);    }    return 0;}

debug的代码（未修改）

 /************************************************************************* > File Name: poj3126.cpp    > Author:LaPucelle    > Mail:     > Created Time: 2017年01月21日 星期六 23时41分38秒 ************************************************************************/#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstdlib>#include<climits>#include<string>#include<cstring>#include<vector>#include<set>#include<list>#include<map>#include<queue>#define rep(i,a,b) for(int i=(a);i<=(b);(i++))#define inf 0x3fff3f#define ll long long#define pi acos(-1)int dire[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };int dire2[8][2]={{-1,-1},{-1,0},{-1,1},{ 0,-1},{ 0,1},{ 1,-1},{ 1,0},{ 1,1}};int dire3[6][3]={ {0,0,1},{0,1,0},{1,0,0},{0,0,-1},{0,-1,0},{-1,0,0} };using namespace std;const int maxn=10000;int prime[maxn];int vis[maxn];inline int isprime(int n){    int now=0;    for(int i=2;i*i<=n;i++){        if(n%i==0){            now=1;            break;        }    }    if(now)        return 0;    return 1;}int gene(int l,int r){    for(int i=l;i<=r;i++){        if(isprime(i)){            prime[i]=1;        }    }}int step[maxn];void bfs(int n,int m){    queue<int> q;    q.push(n);    step[n]=0;    vis[n]=1;    int now=0;    while(!q.empty() ){        int head=q.front();        q.pop();        //printf("@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@\nnow=%d\n",now);        //printf("head=%d\n",head);        /*        if(head==1033){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",1033);        }        if(head==1733){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",1733);        }        if(head==3733){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",3733);        }        if(head==3739){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",3739);        }        if(head==3779){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",3779);        }        if(head==8779){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",8779);        }        if(head==8179){            printf("now=%d\n",now);            printf("step=%d\n",step[head]);            printf("%d\n",8179);        }        */        if(head==m){            //printf("hello world\n");            printf("%d\n",step[m]);            return;        }        int num[5];        num[1]=head/1000;        num[2]=head/100%10;        num[3]=head/10%10;        num[4]=head%10;        /*        for(int i=1;i<=4;i++)            printf("%d ",num[i])            */        int next;        for(int i=4;i>=1;i--){            int j;            if(i==1)                j=1;            else                j=0;            int tmp=num[i];            for(;j<=9;j++){                num[i]=j;                next=num[1]*1000+num[2]*100+num[3]*10+num[4];                if( prime[next] && !vis[next]){                    //printf("next=%d",next);                   /*                     if(next==m){                        printf("nico nico ni\n");                        for(int k=1;k<=4;k++)                            printf("%d",num[k]);                        printf("\n");                        //printf("%d",step[head]+1);                    }                    */                    vis[next]=1;                    q.push(next);                    step[next]=step[head]+1;                }            }            num[i]=tmp;        }                now++;     }}int main(){    //std::ios::sync_with_stdio(false);    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    memset(prime,0,sizeof(prime));    gene(1000,9999);    /*    printf("prime[1033]=%d\n",prime[1033]);    printf("prime[8179]=%d\n",prime[8179]);    */    /*    for(int i=1000;i<=9999;i++)        if(prime[i]==1)            printf("%d\n",i);            */    int t;    cin>>t;    int count=1;    while(t--){        //printf("case %d\n",count++);        int n,m;        cin>>n>>m;        memset(vis,0,sizeof(vis));        memset(step,0,sizeof(step));        bfs(n,m);    }    return 0;}