POJ3126 Prime Path —— BFS + 素数表

题目链接：http://poj.org/problem?id=3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22936		Accepted: 12706

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

题解：

1.打印素数表。

2.由于只有四位数，直接枚举不会超时。由于要求的是“最少步数”，所以用BFS进行搜索。

写法一:

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 struct node //num为素数，dig[]为这个素数的每个位上的数，便于操作。
 {
     int num, dig[], step;
 };

 int vis[], pri[];

 queue<node>que;
 int bfs(node s, node e)
 {
     ms(vis,);
     while(!que.empty()) que.pop();
     s.step = ;
     vis[s.num] = ;
     que.push(s);

     node now, tmp;
     while(!que.empty())
     {
         now = que.front();
         que.pop();

         if(now.num==e.num)
             return now.step;

         for(int i = ; i<; i++)    //枚举位数
         for(int j = ; j<; j++)   //枚举数字
         {
             if(i== && j==) continue;  //首位不能为0
             tmp = now;
             tmp.dig[i] = j;     //第i为变为j
             tmp.num = tmp.dig[] + tmp.dig[]*+tmp.dig[]*+tmp.dig[]*;
             if(!pri[tmp.num] && !vis[tmp.num])  //num为素数并且没有被访问
             {
                 vis[tmp.num] = ;
                 tmp.step++;
                 que.push(tmp);
             }
         }
     }
     return -;
 }

 void init()     //素数表，pri[]==0的为素数
 {
     int m = sqrt(+0.5);
     ms(pri,);
     pri[] = ;
     for(int i = ; i<=m; i++) if(!pri[i])
         for(int j = i*i; j<=; j += i)
             pri[j] = ;
 }

 int main()
 {
     init();
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int n, m;
         node s, e;
         scanf("%d%d",&n,&m);
         s.num = n; e.num = m;
         for(int i = ; i<; i++, n /= ) s.dig[i] = n%;
         for(int i = ; i<; i++, m /= ) e.dig[i] = m%;

         int ans = bfs(s,e);
         if(ans==-)
             puts("Impossible");
         else
             printf("%d\n",ans);
     }
 }

写法二：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 struct node
 {
     int num, step;
 };

 int vis[], dig[];
 int pri[];

 queue<node>que;
 int bfs(int s, int e)
 {
     ms(vis,);
     while(!que.empty()) que.pop();

     node now, tmp;
     now.num = s;
     now.step = ;
     vis[now.num] = ;
     que.push(now);

     while(!que.empty())
     {
         now = que.front();
         que.pop();

         if(now.num==e)
             return now.step;

         dig[] = now.num%;
         dig[] = (now.num/)%;
         dig[] = (now.num/)%;
         dig[] = (now.num/)%;
         for(int i = ; i<; i++)
         for(int j = ; j<; j++)
         {
             if(i== && j==) continue;
             tmp.num = now.num + (j- dig[i])*pow(,i);  //pow前面不要加上强制类型(int)
 //            tmp.num = now.num + (j- dig[i])* (int)pow(10,i);
 //            (int)pow(10,2) 居然等于99， 看来还是不要依赖这些函数
             if(!pri[tmp.num] && !vis[tmp.num])
             {
                 vis[tmp.num] = ;
                 tmp.step = now.step + ;
                 que.push(tmp);
             }
         }
     }
     return -;
 }

 void init()
 {
     int m = sqrt(+0.5);
     ms(pri,);
     pri[] = ;
     for(int i = ; i<=m; i++) if(!pri[i])
         for(int j = i*i; j<=; j += i)
             pri[j] = ;
 }

 int main()
 {
     init();
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int n, m;
         scanf("%d%d",&n,&m);
         int ans = bfs(n,m);
         if(ans==-)
             puts("Impossible");
         else
             printf("%d\n",ans);
     }
 }