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 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
题意:给两个质数n,m,每次只能改变n中的一个数字且改变后的数字依旧是一个质数,问最少需要多少步可以将n变成m。
方法:此题需要先打出一个素数表,然后在广搜,分别将n中的每一位遍历即可。
AC代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
bool vis[maxn];
int prime[maxn];
int v1[maxn];
int plen;
void getp()
{
for(int i=2;i<=10000;i++)
{
if(!v1[i])
{
prime[plen]=i;
plen++;
v1[i]=1;
int q=i;
while(q<=10000)
{
v1[q]=1;
q=q+i;
}
}
}
for(int i=0;i<plen;i++)
{
if(prime[i]>1000)
vis[prime[i]]=true;
}
}
bool vi[maxn];
int u,v;
struct node
{
int x;
int step;
}pn[maxn];
int bfs(int x)
{
queue<node >q;
pn[0].x=x;
pn[0].step=0;
q.push(pn[0]);
vi[x]=true;
while(!q.empty())
{
node front=q.front();
node tail;
q.pop();
int num[4];
int t=0;
while(front.x)
{
num[t]=front.x%10;
front.x=front.x/10;
t++;
}
int nape;
for(int i=1;i<=9;i++)
{
nape=i*1000+num[2]*100+num[1]*10+num[0];
if(vi[nape]==false&&vis[nape]==true)
{
tail.x=nape;
tail.step=front.step+1;
vi[nape]=true;
q.push(tail);
if(tail.x==v)
return tail.step;
}
}
for(int i=0;i<=9;i++)
{
nape=i*100+num[3]*1000+num[1]*10+num[0];
if(vi[nape]==false&&vis[nape]==true)
{
tail.x=nape;
tail.step=front.step+1;
vi[nape]=true;
q.push(tail);
if(tail.x==v)
return tail.step;
}
}
for(int i=0;i<=9;i++)
{
nape=i*10+num[3]*1000+num[2]*100+num[0];
if(vi[nape]==false&&vis[nape]==true)
{
tail.x=nape;
tail.step=front.step+1;
vi[nape]=true;
q.push(tail);
if(tail.x==v)
return tail.step;
}
}
for(int i=0;i<=9;i++)
{
nape=i+num[3]*1000+num[2]*100+num[1]*10;
if(vi[nape]==false&&vis[nape]==true)
{
tail.x=nape;
tail.step=front.step+1;
vi[nape]=true;
q.push(tail);
if(tail.x==v)
return tail.step;
}
}
}
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ios::sync_with_stdio(0),cin.tie(0);
getp();
int n;
scanf("%d",&n);
while(n--)
{
scanf("%d %d",&u,&v);
ms(vi);
memset(pn,0,sizeof(pn));
if(u==v)
printf("0\n");
else
printf("%d\n",bfs(u));
}
return 0;
}