意甲冠军  给你两个4位质数a, b  每次你可以改变a个位数,但仍然需要素数的变化  乞讨a有多少次的能力,至少修改成b

基础的bfs  注意数的处理即可了  出队一个数  然后入队全部能够由这个素数经过一次改变而来的素数  知道得到b

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 10000;
int p[N], v[N], d[N], q[N], a, b; void initPrime()
{
memset(v, 0 , sizeof(v));
for(int i = 2; i * i < N; ++i)
if(!v[i]) for(int j = i; i * j < N; ++j) v[i * j] = 1;
for(int i = 2; i < N ; ++i) p[i] = !v[i];
} int bfs()
{
int c, t, le = 0, ri = 0;
memset(v, 0, sizeof(v));
q[ri++] = a, v[a] = 1, d[a] = 0;
while(le < ri)
{
c = q[le++];
if( c == b) return d[c];
for(int i = 1; i < N; i *= 10)
{
for(int j = 0; j < 10; ++j) //把c第i数量级的数改为j
{
if(i == 1000 && j == 0) continue;
t = c / (i * 10) * i * 10 + i * j + c % i;
if(p[t] && !v[t])
v[t] = 1, d[t] = d[c] + 1, q[ri++] = t;
}
}
}
return -1;
} int main()
{
int cas;
scanf("%d", &cas);
initPrime();
while(cas--)
{
scanf("%d%d", &a, &b);
if((a = bfs()) != -1) printf("%d\n", a);
else puts("Impossible");
}
return 0;
}

Prime Path

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

版权声明:本文博主原创文章,博客,未经同意不得转载。

POJ 3126 Prime Path(BFS 数字处理)的更多相关文章

  1. poj 3126 Prime Path bfs

    题目链接:http://poj.org/problem?id=3126 Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submi ...

  2. poj 3126 Prime Path( bfs + 素数)

    题目:http://poj.org/problem?id=3126 题意:给定两个四位数,求从前一个数变到后一个数最少需要几步,改变的原则是每次只能改变某一位上的一个数,而且每次改变得到的必须是一个素 ...

  3. POJ 3126 Prime Path bfs, 水题 难度:0

    题目 http://poj.org/problem?id=3126 题意 多组数据,每组数据有一个起点四位数s, 要变为终点四位数e, 此处s和e都是大于1000的质数,现在要找一个最短的路径把s变为 ...

  4. POJ 3126 Prime Path(BFS求“最短路”)

    题意:给出两个四位数的素数,按如下规则变换,使得将第一位数变换成第二位数的花费最少,输出最少值,否则输出0. 每次只能变换四位数的其中一位数,使得变换后的数也为素数,每次变换都需要1英镑(即使换上的数 ...

  5. POJ 3126 Prime Path BFS搜索

    题意:就是找最短的四位数素数路径 分析:然后BFS随便搜一下,复杂度最多是所有的四位素数的个数 #include<cstdio> #include<algorithm> #in ...

  6. POJ 3126 Prime Path (BFS+剪枝)

    题目链接:传送门 题意: 给定两个四位数a.b,每次能够改变a的随意一位.而且确保改变后的a是一个素数. 问最少经过多少次改变a能够变成b. 分析: BFS,每次枚举改变的数,有一个剪枝,就是假设这个 ...

  7. POJ 3126 Prime Path (BFS + 素数筛)

    链接 : Here! 思路 : 素数表 + BFS, 对于每个数字来说, 有四个替换位置, 每个替换位置有10种方案(对于最高位只有9种), 因此直接用 BFS 搜索目标状态即可. 搜索的空间也不大. ...

  8. BFS POJ 3126 Prime Path

    题目传送门 /* 题意:从一个数到另外一个数,每次改变一个数字,且每次是素数 BFS:先预处理1000到9999的素数,简单BFS一下.我没输出Impossible都AC,数据有点弱 */ /**** ...

  9. 双向广搜 POJ 3126 Prime Path

      POJ 3126  Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16204   Accepted ...

随机推荐

  1. IIS设置允许下载.exe文件解决方法

    最近很多客户使用IIS服务器,然后提示返现宝下载无法找到等无法下载的问题. 返现宝是.exe安装文件,部分服务器或主机可能无法下载. 第一.如果是自己服务器或VPS请按如下设置: 1.设置MIME,让 ...

  2. .NET开源 FAQ

    Microsoft至2014年11月12日本(PST)公布.NET开源.一个"隐居"商业帝国也迎来"改革开放".. . Q1:为什么要开放源码? Ans:由于. ...

  3. 文件下载-SpringMVC中測试

    直接改动文件路径就能够.其它都不须要改动,帮助类已经为大家写好,可直接使用 1.Scroller: /** * 下载文件 * @author liupeng * @param request * @p ...

  4. 搭建ganglia集群而且监视hadoop CDH4.6

    前言 近期在研究云监控的相关工具,感觉ganglia颇有亮点,能从一个集群总体的角度来展现数据. 但是安装过程稍过复杂,相关依赖稍多,故写此文章与大家分享下. 本文不解说相关原理,若想了解请參考其它资 ...

  5. java 配置及安装Eclipse

    jdk下载 点我~ Java SE Development Kit 8u20 You must accept the Oracle Binary Code License Agreement for ...

  6. swift学习一:介绍,开发文档下载

    在今天wwdc2014公布会上.苹果今天公布了全新的编程语言Swift以及新版Xcode.对于开发人员来说,Swift包括了非常多开发人员喜欢的功能,能够与Objective-C和C语言共同工作.Sw ...

  7. Qt数据类型转换

    把QString转换为double类型 方法1.QString str="123.45"; double val=str.toDouble(); //val=123.45 方法2. ...

  8. BZOJ 3172([Tjoi2013]单词-后缀数组第一题+RMQ)

    3172: [Tjoi2013]单词 Time Limit: 10 Sec   Memory Limit: 512 MB Submit: 268   Solved: 145 [ Submit][ St ...

  9. kb3035583

    dism /online /Get-Packages /Format:Table|findstr 3035583 升级到w10补丁

  10. Gitblit配置

    Gitblit的安装配置及访问-windows (2013-09-11 11:52:31) 转载▼   分类: android基础 Git 是现在很流行的分布式版本控制工具,github更是人人皆知. ...