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 题意:给你两个四位数m和n,求m变到n至少需要几步;每次只能从个十百千上改变一位数字,并且改变后的数要是素数;
我们可以用优先对列做;
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#define N 11000
using namespace std;
int prime(int n)
{
int i,k;
k=(int)sqrt(n);
for(i=;i<=k;i++)
if(n%i==)
return ;
return ;
}
struct node
{
int x,step;
friend bool operator<(node a,node b)
{
return a.step>b.step;
}
};
int bfs(int m,int n)
{
int a[]={,,,};
int vis[N];
priority_queue<node>Q;
node q,p;
memset(vis,,sizeof(vis));
vis[m]=;
q.x=m;
q.step=;
Q.push(q);
while(!Q.empty())
{
q=Q.top();
Q.pop();
if(q.x==n)
return q.step;
for(int i=;i<;i++)//控制改变的哪一位
{
for(int j=;j<;j++)//把相对应的那一位变成j;
{
int L=q.x/(a[i]*);
int R=q.x%(a[i]);
p.x=L*(a[i]*)+j*a[i]+R;//重新组成的数;
if(p.x>=&&vis[p.x]==&&prime(p.x)==)
{
vis[p.x]=;
p.step=q.step+;
Q.push(p);
}
}
}
}
return -;
} int main()
{
int T,m,n,ans;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
ans=bfs(m,n);
if(ans==-)
printf("Impossible\n");
else
printf("%d\n",ans);
}
return ;
}

Prime Path--POJ3126(bfs)的更多相关文章

  1. 素数路径Prime Path POJ-3126 素数,BFS

    题目链接:Prime Path 题目大意 从一个四位素数m开始,每次只允许变动一位数字使其变成另一个四位素数.求最终把m变成n所需的最少次数. 思路 BFS.搜索的时候,最低位为0,2,4,6,8可以 ...

  2. 【POJ - 3126】Prime Path(bfs)

    Prime Path 原文是English 这里直接上中文了 Descriptions: 给你两个四位的素数a,b.a可以改变某一位上的数字变成c,但只有当c也是四位的素数时才能进行这种改变.请你计算 ...

  3. Prime Path(BFS)

    Prime Path Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total S ...

  4. (简单) POJ 3126 Prime Path,BFS。

    Description The ministers of the cabinet were quite upset by the message from the Chief of Security ...

  5. HDU - 1973 - Prime Path (BFS)

    Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total ...

  6. POJ - 3126 - Prime Path(BFS)

    Prime Path POJ - 3126 题意: 给出两个四位素数 a , b.然后从a开始,每次可以改变四位中的一位数字,变成 c,c 可以接着变,直到变成b为止.要求 c 必须是素数.求变换次数 ...

  7. Prime Path POJ-3126

    The ministers of the cabinet were quite upset by the message from the Chief of Security stating that ...

  8. poj3216 Prime Path(BFS)

    题目传送门  Prime Path The ministers of the cabinet were quite upset by the message from the Chief of Sec ...

  9. POJ 3126 Prime Path (bfs+欧拉线性素数筛)

    Description The ministers of the cabinet were quite upset by the message from the Chief of Security ...

  10. Prime Path[POJ3126] [SPFA/BFS]

    描述 孤单的zydsg又一次孤单的度过了520,不过下一次不会再这样了.zydsg要做些改变,他想去和素数小姐姐约会. 所有的路口都被标号为了一个4位素数,zydsg现在的位置和素数小姐姐的家也是这样 ...

随机推荐

  1. Ansible 安装和管理服务

    ansible 使用 yum 模块来安装软件包,使用 service 模块来启动软件: [root@localhost ~]$ ansible 192.168.119.134 -m yum -a &q ...

  2. iOS - App Extension 整体总结

    一.App Extension的介绍 App Extension可以让你扩展你APP的自定义功能和内容,使用户可以在与其他应用或者系统进行互动的时候去使用它.app extension即为本文所说的e ...

  3. xp下对dinput8.dll 游戏键盘输入的模拟 非函数hook

    https://www.xuebuyuan.com/833929.html 很多游戏或者3d模拟软件为了更好的支持外设使用directinput作为输入接口调用.那么如果要模拟鼠标或键盘来控制游戏或者 ...

  4. Matlab 曲线拟合之polyfit与polyval函数

    p=polyfit(x,y,n) [p,s]= polyfit(x,y,n) 说明:x,y为数据点,n为多项式阶数,返回p为幂次从高到低的多项式系数向量p.x必须是单调的.矩阵s用于生成预测值的误差估 ...

  5. open-falcon之alarm、sender、links说明.md

    alarm 功能 处理judge 产生的告警event 区分告警优先级,优先处理级别比较高的告警 为用户提供回调接口 生成告警msg 展示未恢复的告警 配置文件 { "debug" ...

  6. IOS设计模式第六篇之适配器设计模式

    版权声明:原创作品,谢绝转载!否则将追究法律责任. 那么怎么使用适配器设计模式呢? 这个之前提到的水平滚动的视图像这样: 为了开始实现他,我们创建一个新的继承与UIView的HorizontalScr ...

  7. mysql架构图

    整体架构图 访问控制图

  8. linux下MySQL安装及设置(二)

    MySQL二进制分发包安装 去MySql官网下MySQL classic版mysql-5.6.30-osx10.11-x86_64.tar.gz  http://dev.mysql.com/downl ...

  9. 跟bWAPP学WEB安全(PHP代码)--HTML注入和iFrame注入

    背景 这里讲解HTML注入和iFrame注入,其他的本质都是HTML的改变.那么有人会问,XSS与HTML注入有啥区别呢?其实本质上都是没有区别的,改变前端代码,来攻击客户端,但是XSS可以理解为注入 ...

  10. dig命令安装

    yum -y install bind-utils  Dig是一个在类Unix命令行模式下查询DNS包括NS记录,A记录,MX记录等相关信息的工具 查找yahoo.com的A记录:(此处一定是域而不是 ...