Prime Path DescriptionThe 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 e…

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,后来搜了一下解题报告,才发现是bfs(). 想想也是,题目其实已经提示的很清楚了,求最短的路径,对于每一个数,每次可以把4位中的任意一位,  换成与该位不相同的0-9中的任意一位,对于迷宫类 bfs每次转移数为上下左右四个状态,而此题就相当于每次可以转移40个状态(其实最低位为偶数可以排除,不过题目数据…

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) Problem Description The ministers of the cabinet were quite upset by the message from the Chief of Secu…

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 initP…

题目链接:http://poj.org/problem?id=3126 Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 32036   Accepted: 17373 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they…

题目链接:http://poj.org/problem?id=3126 Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22936   Accepted: 12706 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they…

[POJ]P3126 Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35230   Accepted: 18966 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…

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9982   Accepted: 5724 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-digi…

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(即使换上的数是以前被换下的). 思路:若素数a可以按上述规则转化为b,则可以看做a.b直接有一条边.显然,从初始值到目标值的路径上的边数即为花费的 数目,这样一来,就相当于求最短路径.由于题目只要求最小花费数,所以不需要存储有向图. 用BFS搜索,每次枚举当前值x所能变换得到的值y,若y满足条件,将y以及从初始…

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<…