POJ 3126 Prime Path

注意结构体队列的使用，在队列中，一条完整的路径才是正确的，中间的歧路最后还是走不通

转载：http://blog.csdn.net/yulanarti/article/details/1787971  ——> 说明sprintf()的作用

注意 sprintf(num, "%d", cur.num);//将数字转化为字符串，可以替代 itoa

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#define sc(x,y) scanf("%d %d",&(x),&(y))
#define pf(x) printf("%d
", x)
#define CL(x, y) memset(x,y,sizeof(x))
using namespace std;
const int MAX = 100000;
int prime[MAX] = {0};   //素数表
int N, a, b, i, j, temp;
int used[MAX];
void InitPrime();
void BFS(int x, int y);
struct node
{
    int num;
    int step;
};
queue <node> Q;
int main()
{
    InitPrime();
    cin >> N;
    while(N--)
    {
        cin >> a >> b;
        CL(used, 0);
        BFS(a, b);
    }
    return 0;
}
void BFS(int front, int rear)
{
    while(!Q.empty())
        Q.pop();//Q.clear();清空
    node first, cur, next;
    first.num = front;
    first.step = 0;
    used[front] = 1;
    Q.push(first);
    while(!Q.empty())
    {
        cur = Q.front();
        Q.pop();
        if(cur.num == rear)
        {
            cout << cur.step << endl;
            return ;
        }
        for(i = 0; i < 4; i++)
        {
            char num[5];
            sprintf(num, "%d", cur.num);//将数字转化为字符串，可以替代 itoa
            for (j = 0; j < 10; j++)
            {
                if (j == 0 && i == 0)//千位不能为 0
                    continue;
                if(i == 0)
                    temp = j*1000+(num[1]-'0')*100+(num[2]-'0')*10+(num[3]-'0');//千位
                else if(i == 1)
                    temp = j*100+(num[0]-'0')*1000+(num[2]-'0')*10+(num[3]-'0');//百位
                else if(i == 2)
                    temp = j*10+(num[0]-'0')*1000+(num[1]-'0')*100+(num[3]-'0');//十位
                else if(i == 3)
                    temp = j+(num[0]-'0')*1000+(num[1]-'0')*100+(num[2]-'0')*10;//个位
                if(prime[temp] && !used[temp])
                {
                    next.num = temp;
                    next.step = cur.step+1;
                    used[temp] = 1;
//                    cout << next.num << " " << next.step << endl;
                    Q.push(next);
                    Q.push(next);
                }
            }
        }
    }
    printf("Impossible
");
    return ;
}
void InitPrime()
{
    int i, j;
    for(i = 1009; i <= 9973; i++)
    {
        if(!(i&1))continue;  //奇偶判断
        for(j = 3; j*j <= i; j += 2)
        {
            if(0 == i%j)
            {
                break;
            }
        }
        if(j * j > i)
        {
            prime[i] = 1;
        }
    }
}