Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15325   Accepted: 8634 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-dig…

Language: Default Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11703   Accepted: 6640 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to c…

题意:给定两个四位素数作为终点和起点,每次可以改变起点数的某一位,且改变后的数仍然是素数,问是否可能变换成终点数字? 思路:bfs搜索,每次改变四位数中的某一位.素数打表方便判断新生成的数是否是素数. AC代码 #include<cstdio> #include<cstring> #include<queue> #include<cmath> using namespace std; const int maxn = 1e5 + 5; int vis[max…

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14539   Accepted: 8196 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-dig…

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.求变换次数的最小值.(a,b,c都是四位数字,输入时没有前导零) 分析: 每次改变可以获得一个四位数c,然后如果c是素数并且之前没有出现过,那么我们把它放入队列即可. int f[10001]; int v[10001]; void init()//素数筛 { memset(f,0,sizeof f); fo…

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 */ /************************************************ Author :Running_Time Created Time :2015-8-2 15:46:57 File Name :POJ_3126.cpp ****************************…

Prime Path(POJ - 3126) 题目链接 算法 BFS+筛素数打表 1.题目主要就是给定你两个四位数的质数a,b,让你计算从a变到b共最小需要多少步.要求每次只能变1位,并且变1位后仍然为质数. 2.四位数的范围是1000~9999,之间共有1000多个质数.由于已经知道位数为4位,所以可以通过BFS来寻找最小步数.每次需要分别变换个位.十位.百位.千位,并且把符合要求的数放到队列中,同时需标记这个数已经遍历过一次,避免重复遍历,直到找到目标数. C++代码 #include<io…

  POJ 3126  Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16204   Accepted: 9153 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…

POJ 3126 Prime Path(素数路径) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 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…

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数: #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<…