[HDU 1973]--Prime Path(BFS,素数表)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1973
Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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.
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).
the minimal cost or containing the word Impossible.
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
struct node{
int cur, step;
}now, Next;
int vis[], star, finish, prime[] = { , , };
void init(){
for (int i = ; i < ; i++){
if (!prime[i]){
for (int j = ; i*j < ; j++)
prime[i*j] = ;
}
}
}
int bfs(){
queue<node> Q;
vis[star] = ;
now.cur = star, now.step = ;
Q.push(now);
while (!Q.empty()){
int i, j;
char num[];
now = Q.front();
Q.pop();
if (now.cur == finish) return now.step;
for (i = ; i < ; i++){
sprintf(num, "%d", now.cur);
for (j = ; j < ; j++){
if (j == && i == )
continue;
if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j + (num[] - '') * + (num[] - '') * + (num[] - '') * ;
if (!prime[Next.cur] && !vis[Next.cur])
{
Next.step = now.step + ;
vis[Next.cur] = ;
Q.push(Next);
}
}
}
}
return -;
}
int main(){
int t, ans;
cin >> t;
init();
while (t--){
cin >> star >> finish;
memset(vis, , sizeof(vis));
ans = bfs();
if (ans == -) cout << "Impossible\n";
else cout << ans << endl;
}
return ;
}
其实这道题学校OJ(Swust OJ)也有但是坑爹的后台数据变成a+b的后台数据了,Orz~~~(无爱了)
[HDU 1973]--Prime Path(BFS,素数表)的更多相关文章
- hdu - 1195 Open the Lock (bfs) && hdu 1973 Prime Path (bfs)
http://acm.hdu.edu.cn/showproblem.php?pid=1195 这道题虽然只是从四个数到四个数,但是状态很多,开始一直不知道怎么下手,关键就是如何划分这些状态,确保每一个 ...
- hdu 1973 Prime Path
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Description The ministers of the cabi ...
- HDU - 1973 - Prime Path (BFS)
Prime Path Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- POJ3126 Prime Path (bfs+素数判断)
POJ3126 Prime Path 一开始想通过终点值双向查找,从最高位开始依次递减或递增,每次找到最接近终点值的素数,后来发现这样找,即使找到,也可能不是最短路径, 而且代码实现起来特别麻烦,后来 ...
- 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 ...
- 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. ...
- 【BFS】hdu 1973 Prime Path
题目描述: http://poj.org/problem?id=3414 中文大意: 使用两个锅,盛取定量水. 两个锅的容量和目标水量由用户输入. 允许的操作有:灌满锅.倒光锅内的水.一个锅中的水倒入 ...
- POJ 3126 Prime Path(素数路径)
POJ 3126 Prime Path(素数路径) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 The minister ...
- POJ 3216 Prime Path(打表+bfs)
Prime Path Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27132 Accepted: 14861 Desc ...
随机推荐
- C#中网站根路径、应用根路径、物理路径、绝对路径,虚拟路径的区别
C#中网站根路径,请站点的最外一层 /表示 应用根路径 ~/表示,有时候C#程序路径并不是网站路径 物理路径 server.mappath("~/") 是指应用程序放在服务器硬盘的 ...
- 关于php支持的协议与封装协议
<?php /* * php://stdin 标准输入流 * php://stdout 标准输入流 * php://stderr 标准错误流 * php://output 只写的数据流 * ph ...
- 5.4.3 RegExp构造函数属性
RegExp构造函数包含一些属性(这些属性在其他语言中被看成是静态属性).这些属性适用于作用域中的所有正则表达式,并且基于所执行的最近一次正则表达式操作而变化.关于这些属性的另一个独特之处,就是可以通 ...
- ASP.NET MVC Framework
ASP.NET MVC Framework是微软在ASP.NET中所添加的一组类库,这组类库可以使用Model-View-Controller的设计模式来开发ASP.NET的应用程序.它与现有的ASP ...
- A Byte of Python 笔记(7)数据结构:列表、元组、字典,序列
第9章 数据结构 数据结构,即可以处理一些数据的结构.或者说,它们是用来存储一组相关数据的. python 有三种内建的数据结构--列表.元组和字典. list = ['item1', 'item2' ...
- 数据库导出到excel
项目结构同上一篇 泛型通用的写法 ExportExcel.java package excel; import java.io.OutputStream; import java.lang.refle ...
- C# 读书笔记之继承与多态
1.1继承与多态的基本概念 1.1.1 继承和多态 继承是面向对象程序设计的主要特征之一,允许重用现有类(基类,亦称超类.父类)去创建新类(子类,亦称派生类)的过程.子类将获取基类的所有非私有数据和行 ...
- perl5 第十二章 Perl5中的引用/指针
第十二章 Perl5中的引用/指针 by flamephoenix 一.引用简介二.使用引用三.使用反斜线(\)操作符四.引用和数组五.多维数组六.子程序的引用 子程序模板七.数组与子程序八.文件句 ...
- Java ThreadLocal 学习
同步机制是为了同步多个线程对相同资源的并发访问,是为了多个线程之间进行通信的有效方式. 而ThreadLocal是隔离多个线程的数据共享,从根本上就不在多个线程之间共享资源(变量),这样当然不需要对多 ...
- java学习之即时通信项目实战
项目总结:这次项目主要是根据视频来的,结果跟到一半感觉跟不上,慢慢自己有了自己的想法,决定自己先不看学习视频,自己先试着写. 总结写前面,算是写的第一个项目吧.项目中遇到几点问题,首先Scoket对 ...