BFS (Prime Path , POJ 3126)

基本都是一样的套路！！！！！！！！所以加深熟练度！！！！！！！！

部分题目如下：所有数都要求为素数，每次替换某一位数，得到目标数。

— 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 1e4+5;
const int INF = 0x3f3f3f3f;
int is_prime[maxn];
int vis[maxn];
char chage[10] = {'0','1','2','3','4','5','6','7','8','9'};
struct node
{
    char anum[10];
    int step;
}s,e;
queue<node> que;
void prime() //1000-10000素数打表
{
    is_prime[1] = 1;
    for(int i = 2; i < maxn; i++)
    {
        if(!is_prime[i])
            for(int j = 2*i; j < maxn; j+=i)
                is_prime[j] = 1;
    }
}
int getnum(char *a) //字符数组转成数
{
    int temp = 0;
    for(int i = 0; i < strlen(a); i++)
        temp = temp*10+a[i]-'0';
    return temp;
}
void bfs()
{
    queue<node> que;
    s.step = 0;
    que.push(s);
    vis[getnum( s.anum) ] = 1; //经过的点标记，避免个回头
    node next,temp;
    while(!que.empty())
    {
        temp = que.front();
        //cout<<temp.anum<<endl;
        que.pop();
        if(getnum(temp.anum) == getnum(e.anum))
        {
            cout<<temp.step<<endl;
            return ;
        }
        for(int i = 0; i < 10; i++)
        {
            for(int j = 0; j < 4; j++)
            {
                next = temp;
                next.anum[j] = chage[i];
                int item = getnum( next.anum);
                if(!vis[item] && !is_prime[item] && item > 999 && item < 10000)
                {
                    next.step = temp.step +1;
                    vis[item] = 1;
                    que.push(next);
                }
            }
        }
    }
    cout<<"Impossible"<<endl;
    return ;
}
int main()
{
    memset(is_prime,0,sizeof is_prime); // 等于0表示是素数
    prime();
    int n;
    cin>>n;
    while(n--)
    {
        memset(vis,0,sizeof vis); //等于0表示没经过
        scanf("%s %s",s.anum,e.anum);
        bfs();
    }
    return 0;
}