poj3126(Prime Path)

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstring>
  6 #include <cstdio>
  7 #include <string>
  8 #include <bitset>
  9 #include <vector>
 10 #include <queue>
 11 #include <stack>
 12 #include <cmath>
 13 #include <list>
 14 //#include <map>
 15 #include <set>
 16 using namespace std;
 17 /***************************************/
 18 #define ll long long
 19 #define int64 __int64
 20 #define PI 3.1415927
 21 /***************************************/
 22 const int INF = 0x7f7f7f7f;
 23 const double eps = 1e-8;
 24 const double PIE=acos(-1.0);
 25 const int d1x[]= {0,-1,0,1};
 26 const int d1y[]= {-1,0,1,0};
 27 const int d2x[]= {0,-1,0,1};
 28 const int d2y[]= {1,0,-1,0};
 29 const int fx[]= {-1,-1,-1,0,0,1,1,1};
 30 const int fy[]= {-1,0,1,-1,1,-1,0,1};
 31 const int dirx[]= {-1,1,-2,2,-2,2,-1,1};
 32 const int diry[]= {-2,-2,-1,-1,1,1,2,2};
 33 /*vector <int>map[N];map[a].push_back(b);int len=map[v].size();*/
 34 /***************************************/
 35 void openfile()
 36 {
 37     freopen("data.in","rb",stdin);
 38     freopen("data.out","wb",stdout);
 39 }
 40 priority_queue<int> qi1;
 41 priority_queue<int, vector<int>, greater<int> >qi2;
 42 /**********************华丽丽的分割线,以上为模板部分*****************/
 43 const int M=10001;
 44 int vis[M],cnt[M];
 45 int isprime[M];
 46 int di[10]= {0,1,2,3,4,5,6,7,8,9};
 47 int ea;
 48 int ce;
 49 void judgeprime()
 50 {
 51     int i,j;
 52     isprime[0]=1;
 53     isprime[1]=1;
 54     for(i=2; i<sqrt(M); i++)
 55         if (!isprime[i])
 56         {
 57             for(j=i+i; j<M; j+=i)
 58                 isprime[j]=1;
 59         }
 60 }
 61 int BFS(int a)
 62 {
 63     queue<int >Q;
 64     int i,j,k,h;
 65     int aa,ge,shi,bai,qi;
 66     Q.push(a);;
 67     while(!Q.empty())
 68     {
 69         int aaa=Q.front();
 70         Q.pop();
 71         vis[aaa]=1;
 72         if (aaa==ea)
 73         {
 74             ce=cnt[aaa];
 75             return 0;
 76         }
 77         int ce1=aaa;
 78         for(i=1; i<10; i+=2) //个位
 79         {
 80             aaa=ce1;
 81             aaa=aaa-aaa%10;
 82             aa=aaa+di[i];
 83             if(!vis[aa]&&!isprime[aa])
 84             {
 85                 cnt[aa]=cnt[ce1]+1;
 86                 Q.push(aa);
 87                 vis[aa]=1;
 88             }
 89         }
 90 
 91         for(i=0; i<10; i++) //十位
 92         {
 93             aaa=ce1;
 94             ge=aaa%10;
 95             aaa=aaa/100;
 96             aa=aaa*100+di[i]*10+ge;
 97             if(!vis[aa]&&!isprime[aa])
 98             {
 99                 cnt[aa]=cnt[ce1]+1;
100                 Q.push(aa);
101                 vis[aa]=1;
102             }
103         }
104 
105         for(i=0; i<10; i++) //百位
106         {
107             aaa=ce1;
108             ge=aaa%10;
109             aaa=aaa/10;
110             shi=aaa%10;
111             aaa=aaa/100;
112             aa=aaa*1000+di[i]*100+shi*10+ge;
113             if(!vis[aa]&&!isprime[aa])
114             {
115                 cnt[aa]=cnt[ce1]+1;
116                 Q.push(aa);
117                 vis[aa]=1;
118             }
119         }
120 
121         for(i=1; i<10; i++) //千位
122         {
123             aaa=ce1;
124             ge=aaa%10;
125             aaa=aaa/10;
126             shi=aaa%10;
127             aaa=aaa/10;
128             bai=aaa%10;
129             aaa=aaa/100;
130             aa=di[i]*1000+bai*100+shi*10+ge;
131             if(!vis[aa]&&!isprime[aa])
132             {
133                 cnt[aa]=cnt[ce1]+1;
134                 Q.push(aa);
135                 vis[aa]=1;
136             }
137         }
138     }
139     return 0;
140 }
141 int main()
142 {
143 
144     int cas;
145     scanf("%d",&cas);
146     while(cas--)
147     {
148         int sa;
149         memset(vis,0,sizeof(vis));
150         memset(cnt,0,sizeof(cnt));
151         memset(isprime,0,sizeof(isprime));
152         ce=-1;
153         scanf("%d%d",&sa,&ea);
154         judgeprime();
155         BFS(sa);
156         if (ce==-1)
157             printf("Impossible\n");
158         else
159             printf("%d\n",ce);
160     }
161     return 0;
162 }