文档章节

poj_3126Prime Path

N3verL4nd
 N3verL4nd
发布于 2017/03/25 10:45
字数 757
阅读 2
收藏 0

Prime Path
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8725   Accepted: 4960

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

筛选法求出素数表,然后BFS,不加visited数组,poj超时,hdoj溢出,毕竟重复的节点还是蛮多的。

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define MAX 9999

typedef struct Node
{
	int x;
	int cnt;
}Node;
int primer[MAX];
bool visited[MAX];
queue<Node>Q;
int y;

void pri() //1代表素数
{
	for(int i = 1; i < MAX; i++)
	{
		primer[i] = 1;
	}
	primer[0] = 0;
	primer[1] = 0;
	for(int i = 2; i < MAX / 2; i++)
	{
		for(int j = 2; i * j < MAX; j++)
		{
			primer[i*j] = 0;
		}
	}
}

int bfs(int x)
{
	int a, b, c;
	while (!Q.empty())
	{
		Q.pop();
	}
	Node pre, next;
	pre.x = x;
	pre.cnt = 0;
	visited[pre.x] = true;
	Q.push(pre);
	while (!Q.empty())
	{
		pre = Q.front();
		Q.pop();
		if(pre.x == y)
		{
			//cout << "YES" << endl;
			return pre.cnt;
		}

		c = pre.x / 10; //个位
		for(int i = 0; i <= 9; i++)
		{
			if(primer[c * 10 + i] == 1 && c * 10 + i != pre.x && !visited[c * 10 + i])
			{
				visited[c * 10 + i] = true;
				next.x = c * 10 + i;
				next.cnt = pre.cnt + 1;
				Q.push(next);
			}
		}

		a = pre.x % 10; //十位
		b = pre.x / 100;
		for(int i = 0; i <= 9; i++)
		{
			if(primer[(b * 10 + i) * 10 + a] == 1 && (b * 10 + i) * 10 + a != pre.x && !visited[(b * 10 + i) * 10 + a])
			{
				visited[(b * 10 + i) * 10 + a] = true;
				next.x = (b * 10 + i) * 10 + a;
				next.cnt = pre.cnt + 1;
				Q.push(next);
			}
		}

		a = pre.x % 100;  //百位
		b = pre.x / 1000;
		for(int i = 0; i <= 9; i++)
		{
			if(primer[100*(b * 10 + i) + a] == 1 && 100 * (b * 10 + i) + a != pre.x && !visited[100 * (b * 10 + i) + a])
			{
				visited[100 * (b * 10 + i) + a] = true;
				next.x = 100 * (b * 10 + i) + a;
				next.cnt = pre.cnt + 1;
				Q.push(next);
			}
		}

		a = pre.x % 1000; //千位
		for(int i = 1; i <= 9; i++)
		{
			if(primer[i * 1000 + a] == 1 && i * 1000 + a != pre.x && !visited[i * 1000 + a])
			{
				visited[i * 1000 + a] = true;
				next.x = i * 1000 + a;
				next.cnt = pre.cnt + 1;
				Q.push(next);
			}
		}
	}
	return 0;
}

int main()
{
	pri();
	int x, t;
	cin >> t;
	while (t--)
	{
		memset(visited, false, sizeof(visited));
		cin >> x >> y;
		cout << bfs(x) << endl;
	}
}


© 著作权归作者所有

下一篇: LoadIcon的使用
N3verL4nd
粉丝 1
博文 379
码字总数 481243
作品 0
朝阳
私信 提问
算法进阶路径

第一阶段:练经典常用算法,下面的每个算法给我打上十到二十遍,同时自己精简代码, 因为太常用,所以要练到写时不用想,10-15分钟内打完,甚至关掉显示器都可以把程序打 出来. 1.最短路(Fl...

暖冰
2016/04/02
155
1
一个搞ACM需要掌握的算法

ACM的竞赛性强,因此自己应该和自己的实际应用联系起来.适合自己的才是好的,有的人不适合搞算法,喜欢系统架构,因此不要看到别人什么就眼红,发挥自己的长处,这才是重要的. 第一阶段:练经典常用...

long0404
2015/06/24
0
0
Tree(树链剖分+线段树延迟标记)

Tree http://poj.org/problem?id=3237 Description You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each ......

Fighting_sh
2018/09/25
0
0
POJ -- 3126 Prime Path

题目网址: POJ -- 3126 给出变化规律,问最少变化几次。使用宽搜就可以解决,每个数的出度是40。 这道题目还用到了简单的素数判定,四位数不大,判断一个四位数是否是素数只要O(100)的复杂度...

傅芃芃
2016/03/11
19
0
POJ -- 2965 The Pilots Brothers' refrigerator

题目链接: POJ -- 2965 题目给出的是4X4的矩阵,规模不算大。由于问的是最少的操作步数,很容易想到bfs宽搜。一共16个格子,总共的状态数是2^16,不算大,所以这个策略的确是对的。 但是需要...

傅芃芃
2016/03/19
27
0

没有更多内容

加载失败,请刷新页面

加载更多

SpringBoot中 集成 redisTemplate 对 Redis 的操作(二)

SpringBoot中 集成 redisTemplate 对 Redis 的操作(二) List 类型的操作 1、 向列表左侧添加数据 Long leftPush = redisTemplate.opsForList().leftPush("name", name); 2、 向列表右......

TcWong
今天
5
0
排序––快速排序(二)

根据排序––快速排序(一)的描述,现准备写一个快速排序的主体框架: 1、首先需要设置一个枢轴元素即setPivot(int i); 2、然后需要与枢轴元素进行比较即int comparePivot(int j); 3、最后...

FAT_mt
昨天
4
0
mysql概览

学习知识,首先要有一个总体的认识。以下为mysql概览 1-架构图 2-Detail csdn |简书 | 头条 | SegmentFault 思否 | 掘金 | 开源中国 |

程序员深夜写bug
昨天
10
0
golang微服务框架go-micro 入门笔记2.2 micro工具之微应用利器micro web

micro web micro 功能非常强大,本文将详细阐述micro web 命令行的功能 阅读本文前你可能需要进行如下知识储备 golang分布式微服务框架go-micro 入门笔记1:搭建go-micro环境, golang微服务框架...

非正式解决方案
昨天
8
0
前端——使用base64编码在页面嵌入图片

因为页面中插入一个图片都要写明图片的路径——相对路径或者绝对路径。而除了具体的网站图片的图片地址,如果是在自己电脑文件夹里的图片,当我们的HTML文件在别人电脑上打开的时候图片则由于...

被毒打的程序猿
昨天
9
0

没有更多内容

加载失败,请刷新页面

加载更多

返回顶部
顶部