打素数表,构造矩阵,然后BFS可得答案

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
int max(int a,int b){return a<b?b:a;}
int min(int a,int b){return a>b?b:a;}
const int maxn=200;
bool p[maxn*maxn*2];
int x[maxn+5][maxn+5];
int y[maxn+5][maxn+5];
bool way[maxn][maxn];
int xx[4]={0,0,1,-1};
int yy[4]={1,-1,0,0};
struct node
{int x,y,dis;node(int a,int b,int c){x=a;y=b;dis=c;}
};
void prime()
{int i=0,j=0,a=maxn,k=a*a;p[1]=1;for(i=2;i<=maxn*maxn;i++){for(j=2*i;j<=maxn*maxn;j+=i){if(p[i]==0)p[j]=1;}}i=-1,j=0;while(k!=0){for(i++;i<a && x[j][i]==0;i++){x[j][i]=k--;}i--;for(j++;j<a && x[j][i]==0;j++){x[j][i]=k--;}j--;for(i--;i>=0 && x[j][i]==0;i--){x[j][i]=k--;}i++;for(j--;j>=0 && x[j][i]==0;j--){x[j][i]=k--;}j++;}for(j=0;j<a;j++)// i 是 x , j 是 y{for(i=0;i<a;i++){if(p[x[j][i]]==1){y[j][i]=1;}}}
}int main()
{prime();int i,j,a,b,t=1;while(cin>>a>>b){memset(way,0,sizeof(way));int sx,sy,ex,ey;for(i=0;i<maxn;i++){for(j=0;j<maxn;j++){if(x[j][i]==a){sx=i;sy=j;}if(x[j][i]==b){ex=i;ey=j;}}}bool flag=true;queue<node> q;while(!q.empty())q.pop();q.push(node(sx,sy,0));way[sy][sx]=true;while(!q.empty())  {  if(q.front().x==ex && q.front().y==ey){flag=false;break;}if(q.front().x >= 0 && q.front().y >=0)  {  for(i=0;i<4;i++)  {  if(y[q.front().y + yy[i] ][q.front().x + xx[i] ] == 1 && way[q.front().y + yy[i] ][q.front().x + xx[i]] == false)  {  way[q.front().y + yy[i] ][q.front().x + xx[i]] = true;  q.push(node( q.front().x + xx[i],q.front().y + yy[i],q.front().dis+1));  }  }  q.pop();  }else q.pop();  }if(flag)cout<<"Case "<<t<<": impossible"<<endl;else cout<<"Case "<<t<<": "<<q.front().dis<<endl;t++;}return 0;
}