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POJ 3126 Prime Path 素数筛+bfs
ac代码： #include <iostream> #include <time.h> #include<stdlib.h> #include<math.h> #include<cstdio> #include<queue> using namespace std; const int maxn=10010; bool vis[maxn]; int prime[maxn/3]; void prime_o(int n) { iPOJ - 3126 Prime Path
题目链接：T_T 题目：Prime Path Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37219 Accepted: 19924*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 thepoj-3126 Prime Path
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 keepPOJ-3126 Prime Path
F - Prime Path (POJ - 3126) using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int N = 1e5+5; bool isp[N] = {0}; bool vis[N] = {0}; int n, s, e; struct P { int x; int step; }; void euler() { int p[N], m=0; f【POJ - 3126】Prime Path（bfs）
Prime Path 原文是English 这里直接上中文了 Descriptions: 给你两个四位的素数a，b。a可以改变某一位上的数字变成c，但只有当c也是四位的素数时才能进行这种改变。请你计算a最少经过多少次上述变换才能变成b。例如：1033 -> 8179 1033 1733 3733 3739 3779 8779 8179最少变换了6