预处理出质数,化成图论问题,dijkstra求最短路
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
#define pii pair<int,int>
#define fi first
#define se second
const int N = 1e4,M =N*2;
int T,p[N],cnt,id[N],S,E,dis[N];
int h[N],e[M],ne[M],idx;
bool st[N];
priority_queue<pii,vector<pii>,greater<pii> > q;
bool check(int a, int b)
{
int num1[4];
for(int i = 0 ; a ; i ++ )
{
num1[i] = a%10;
a/=10;
}
int cnt = 0 ;
for(int i = 0 ; b ; i ++ )
{
if(num1[i] != b % 10) cnt++;
b/=10;
}
if(cnt >= 2 ) return false;
return true;
}
void add(int a,int b)
{
e[idx] = b , ne[idx] = h[a] , h[a] = idx++;
}
void get_primes(int n)
{
for(int i=2;i<=n;i++)
{
if(!st[i])
{
p[cnt]=i;
id[i] = cnt;
cnt++;
}
for(int j=0;p[j]<=n/i;j++)
{
st[p[j]*i]=true;
if(i%p[j]==0)break;//当条件成立 pj就是i的最小质因数
}
}
}
int dijkstra()
{
while(q.size()) q.pop();
memset(dis,0x3f,sizeof dis);
memset(st,0,sizeof st);
q.push({0,S});
dis[S] = 0;
while(q.size())
{
pii t = q.top();
q.pop();
int dist = t.fi,ver = t.se;
if(st[ver]) continue;
st[ver] = true;
for(int i = h[ver] ; ~i ; i = ne[i])
{
int j = e[i];
if(dis[j] > dist + 1 )
{
dis[j] = dist +1;
q.push({dis[j],j});
}
}
}
return dis[E];
}
void sol()
{
int a,b;
cin>>a>>b;
S= id[a] , E = id[b];
int ans = dijkstra();
cout << ans <<endl ;
}
int main()
{
get_primes(9999);
memset(h,-1,sizeof h);
for(int i = id[1009] ; i < cnt ; i ++ )
{
for(int j = i+1 ; j < cnt ; j ++ )
if(check(p[i],p[j]))
{
add(i,j),add(j,i);
//printf("%d & %d is added.\n",p[i],p[j]);
}
}
cin >>T;
while(T--) sol();
return 0;
}