关于素数打表和最短路径(宽度优先搜索)
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
这道题目的意思是给定两个数,求它们之间的最短路径,但是要求路径全部为素数并且每次只能够变换一位数字(狗比的要求)。解决问题的办法是首先要用一个数组来存储所有1000到9999(题目说明为4位数)然后用bfs来寻找这一最短路径。
源代码:
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <cstdlib>
#include <queue>
using namespace std;
#define N 10100
int prime[N];
struct node ///BFS记录步数,习惯用结构体,比较清晰,话说用数组我还不一定写的出来啊....
{
int x,step;
};
void init() ///打N以内的素数表,prime[i]=0为素数。
{
int i,j;
memset(prime,0,sizeof(prime));
prime[1]=1;
for(i=2; i<N; i++)
{
if(prime[i]==0)
{
for(j=2; j*i<N; j++) ///傻逼XX的这里的J是从2开始的。这么简单的地方也会错,唉....
prime[j*i]=1;
}
}
}
int BFS(int s,int e)
{
int vis[N],num; ///vis[N]用来标记是否查找过
memset(vis,0,sizeof(vis));
queue<node>Q; ///申请队列。
node p,q;
p.x=s; ///将起始值赋值给P
p.step=0;
vis[s]=1;
Q.push(p); ///将p压入队列
while(!Q.empty()) ///当队列不为空时
{
p=Q.front(); ///取队首
Q.pop(); ///清队首
if(p.x==e) ///若找到终止值 返回步数,
return p.step;
int t[5];
t[1]=p.x/1000; ///记录千位数
t[2]=p.x/100%10; ///记录百位数
t[3]=p.x/10%10; ///记录十位
t[4]=p.x%10; ///记录各位
for(int i=1;i<=4;i++)
{
int temp=t[i]; ///这里特别注意,每次只能变换一位数,这里用来保存该变换位的数,变换下一位数时,该位数要恢复
for(int j=0;j<10;j++)
{
if(t[i]!=j)
{
t[i]=j; ///t[i]要改变
num=t[1]*1000+t[2]*100+t[3]*10+t[4];
}
if(num>=1000 && num<=9999 && !vis[num] && !prime[num]) ///判断该4位数是否满足条件,
{
q.x=num;
q.step=p.step+1; ///若满足,步数加1,压入队列,标记...
Q.push(q);
vis[num]=1;
}
}
t[i]=temp; ///恢复
}
}
return -1;
}
int main()
{
int n,a,b,ans;
init();
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&a,&b);
ans=BFS(a,b);
if(ans!=-1)
printf("%d\n",ans);
else
printf("Impossible\n");
}
}