POJ——3126Prime Path（双向BFS+素数筛打表）

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16272 Accepted: 9195

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

31033 81791373 80171033 1033

Sample Output

670

单向BFS水题，但是双向让我调试了很久，因为写单向的时候是分4种情况，然后想着双向用for来放在一个循环里好了，结果样例输出答案完全不对，只有答案为1或2的时候可能会对，不解一个早上= =刚才想着算了把for去掉写麻烦点，结果又因为忘记删掉调试输出的语句，WA几发。现在还是不知道为什么原来的for是错的。双向BFS给我一个感觉：代码真长（虽然大部分是重复的）嗯这题写完之后上学期遗留的题除了一道题意本身不清楚+大部分AC代码本身也有明显错误的那道题之外全部A掉了。看看专题训练status里我刷了好几页的历史。真是个悲伤的故事……双向的时候要按层搜索

代码：

#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<cstdio>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define MM(x) memset(x,0,sizeof(x))#define MMINF(x) memset(x,INF,sizeof(x))typedef long long LL;const double PI=acos(-1.0);const int N=10010;int vis[N],prime[N];int color[N];struct info{char s[6];int step;};info S,T;int change(char s[]){int r=0;for (int i=0; i<4; i++)r=r*10+s[i]-'0';return r;}queue<info>Qf,Qb;int T_bfs(){S.step=0;T.step=0;int lay=0;while (!Qf.empty())Qf.pop();while (!Qb.empty())Qb.pop();Qf.push(S);Qb.push(T);color[change(S.s)]=1;color[change(T.s)]=2;while ((!Qf.empty())||(!Qb.empty())){while(!Qf.empty()&&Qf.front().step==lay){info now=Qf.front();Qf.pop();info v=now;while (--v.s[0]>='1'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[0]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (--v.s[1]>='0'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[1]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (--v.s[2]>='0'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[2]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (--v.s[3]>='0'){int num=change(v.s);if(num%2==0)continue;if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[3]<='9'){int num=change(v.s);if(num%2==0)continue;if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=1;vis[num]=v.step;Qf.push(v);}else if(color[num]==2){return vis[num]+vis[change(now.s)];}}}}//while(!Qb.empty()&&Qb.front().step==lay){info now=Qb.front();Qb.pop();info v=now;while (--v.s[0]>='1'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[0]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}v=now;while (--v.s[1]>='0'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[1]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}////v=now;while (--v.s[2]>='0'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}if(color[num]==1){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[2]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}//v=now;while (--v.s[3]>='0'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}v=now;while (++v.s[3]<='9'){int num=change(v.s);if(prime[num]){if(!color[num]){v.step=now.step+1;color[num]=2;vis[num]=v.step;Qb.push(v);}else if(color[num]==1){return vis[num]+vis[change(now.s)];}}}}lay++;}}int main(void){int tcase,i,j;for (i=0; i<N; i++)prime[i]=1;for (i=2; i<N; ++i)for (j=2; j*i<N; ++j)prime[i*j]=0;scanf("%d",&tcase);while (tcase--){MM(vis);MM(color);scanf("%s%s",S.s,T.s);if(strcmp(S.s,T.s)==0)puts("0");elseprintf("%d\n",T_bfs()+1);}return 0;}