B - 小学生
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
8179The 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 1033Sample Output
6 7 0
题意:给你两个四位素数,每次改变其中一位,使得改后依旧是素数,求多少步第一个素数可以变成第二个
爆搜,每次改变一个数字(4*10),搜索是否可行
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <math.h>
typedef long long LL;
typedef long double LD;
using namespace std;
const int maxn=10;
int vis[maxn][maxn][maxn][maxn];
int f[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
int n,m;
struct node
{
int a[5];
int step;
};
queue<node>q;
bool OK(node t,node en)
{
for(int i=1;i<=4;i++)
{
if(t.a[i]!=en.a[i])
{
return false;
}
}
return true;
}
bool ss(node x)
{
int xx=0;
for(int i=1;i<=4;i++)
{
xx=xx*10+x.a[i];
}
for(int i=2;i<=sqrt(xx);i++)
{
if(xx%i==0)return false;
}
return true;
}
void bfs(node st,node en)
{
q.push(st);
while(!q.empty())
{
node t=q.front();
// printf("t:%d%d%d%d %d\n",t.a[1],t.a[2],t.a[3],t.a[4],t.step);
q.pop();
if(OK(t,en))
{
printf("%d\n",t.step);
return;
}
for(int i=1;i<=4;i++)
{
for(int j=1;j<=9;j++)
{
node tem;
tem.a[1]=t.a[1];
tem.a[2]=t.a[2];
tem.a[3]=t.a[3];
tem.a[4]=t.a[4];
tem.step=t.step+1;
tem.a[i]+=j;
tem.a[i]%=10;
if(tem.a[1]==0)continue;
if(!ss(tem)||vis[tem.a[1]][tem.a[2]][tem.a[3]][tem.a[4]]!=0)continue;
q.push(tem);
vis[tem.a[1]][tem.a[2]][tem.a[3]][tem.a[4]]=1;
//printf("tem:%d%d%d%d\n",tem.a[1],tem.a[2],tem.a[3],tem.a[4]);
}
}
}
printf("Impossible\n");
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int x,y;
scanf("%d%d",&x,&y);
memset(vis,0,sizeof(vis));
while(!q.empty())q.pop();
node st,en;
st.step=0;
int i=4;
while(x){st.a[i]=x%10;x/=10;i--;}
vis[st.a[1]][st.a[2]][st.a[3]][st.a[4]]=1;
i=4;
while(y){en.a[i]=y%10;y/=10;i--;}
bfs(st,en);
}
return 0;
}