POJ 3126 Prime Path（素数变换路径）

题目：

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

这个题目是用广度优先搜索。

每次都出队得到一个数，然后对和这个数只隔了一个数码的数（一共有8+9+9+9=35个）进行判断，如果是素数那么就进队。

这样我开始的代码是：

#include<iostream>#include<queue>using namespace std;int n, a, b;int list[10000];queue<int>q;bool is_prime(int n){if (n % 2 == 0)return false;for (int i = 3; i*i <= n; i += 2)if (n%i == 0)return false;return true;}void f(int t){if (is_prime(t) && t != a){q.push(t);list[t] = list[a] + 1;}}int main(){int a1, a2, a3, a4;cin >> n;while (n--){cin >> a >> b;q.push(a);int sum = 0;memset(list, -1, sizeof(list));list[a] = 0;while (!q.empty() && list[b]<0){a = q.front();//a=a1a2a3a4q.pop();a4 = a % 10;a3 = (a / 10) % 10;a2 = (a / 100) % 10;a1 = a / 1000;int t;for (int i = 1; i < 10; i++)f(a - a1 * 1000 + i * 1000);for (int i = 0; i < 10; i++){f(a - a2 * 100 + i * 100);f(a - a3 * 10 + i * 10);f(a - a4 + i);}}cout << list[b] << endl;}return 0;}

运行的结果是

4
1033 8179
1873
1373 8017
55
1033 1033
0

发现结果非常大。

然后我把函数改成了

void f(int t){     if (is_prime(t) && t != a)    {          q.push(t);          if(list[t]<0)list[t] = list[a] + 1;     }}

运行结果是

4
1033 8179
6
1373 8017
4
1033 1033
0

然后再仔细看这个函数，发现还是不对，又改成了

void f(int t){if (is_prime(t) && t != a && list[t] < 0){q.push(t);list[t] = list[a] + 1;}}

在进队之前就判断有没有被访问过。

其实很明显就应该这样，但是以前我对队列不熟，对广度优先也不熟，很多问题都是用深度优先的回溯做的。

运行结果是

4
1033 8179
6
1373 8017
8
1033 1033
0

只有中间的那一组是错的，单独运行

2
1373 8017
7

结果又是对的。

所以我认为是队列没有清空的原因。

加了一句 while(q.size())q.pop(); 之后

代码变成：

#include<iostream>#include<queue>using namespace std;int n, a, b;int list[10000];queue<int>q;bool is_prime(int n)//判断是不是素数{if (n % 2 == 0)return false;for (int i = 3; i*i <= n; i += 2)if (n%i == 0)return false;return true;}void f(int t)//这个函数是用来进队的，写这个函数只是为了节约重复代码{if (is_prime(t) && t != a && list[t] < 0){q.push(t);list[t] = list[a] + 1;}}int main(){int a1, a2, a3, a4;cin >> n;while (n--){cin >> a >> b;while(q.size())q.pop();//清空队列q.push(a);int sum = 0;memset(list, -1, sizeof(list));//初始化listlist[a] = 0;while (!q.empty()&&list[b]<0){a = q.front();//a=(a1a2a3a4)10进制q.pop();a4 = a % 10;a3 = (a / 10) % 10;a2 = (a / 100) % 10;a1 = a / 1000;int t;for (int i = 1; i < 10; i++)f(a - a1 * 1000 + i * 1000);for (int i = 0; i < 10; i++){f(a - a2 * 100 + i * 100);f(a - a3 * 10 + i * 10);f(a - a4 + i);}}cout << list[b] << endl;}return 0;}

然后运行结果对了，一提交，果然是对的。