POJ 3126 BFS

题意

将一个四位的素数，变为另一个四位的素数。每次只能换一位上的数字，不能存在前导0，且中间过程中，换掉一位数字后的数也都是素数。

问最少变换的次数

分析

bfs即可

代码

 1 /* When all else is lost the future still remains. */
 2 /* You can be the greatest */
 3 #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++)
 4 #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--)
 5 #define fi first
 6 #define se second
 7 #define mk(X,Y) make_pair((X),(Y))
 8 #define inf 0x3f3f3f3f
 9 #define clr(X,Y) memset(X,Y,sizeof(X))
10 #define pb push_back
11 //head
12 #include <iostream>
13 #include <stdio.h>
14 #include <queue>
15 #include <algorithm>
16 #include <string>
17 #include <map>
18 #include <string.h>
19 using namespace std;
20 bool primer[10010];
21 int val[10010];
22 bool cur[10010];
23 bool is(int in){
24     for(int i = 2 ; i * i <= in ; i++)
25         if(in % i == 0) return 0;
26     return 1;
27 }
28 void init(){
29     clr(primer,0);
30     rep(i,1000,10000) if(is(i)) primer[i] = 1;
31     return ;
32 }
33 int num(int in[5]){
34     return in[0]*1000+in[1]*100+in[2]*10+in[3];
35 }
36 void slove(int be , int end){
37     clr(val,0);
38     clr(cur,0);
39     //
40     queue<int> Q;
41     Q.push(be);
42     //
43     while(!Q.empty()){
44         if(cur[end]) break;
45         int now = Q.front();
46         cur[now] = 1;
47         int t = now;
48         Q.pop();
49         int p[5];
50         drep(i,3,0){
51             p[i] = t % 10;
52             t /= 10;
53         }
54         rep(i,0,4){
55             int a = p[i];
56             rep(j,0,10){
57                 if(!i && !j) continue;
58                 if(a == j) continue;
59                 p[i] = j;
60                 int k = num(p);
61                 if(!primer[k]) continue;
62                 if(cur[k]) continue;
63                 val[k] = val[now] + 1;
64                 cur[k] = 1;
65                 Q.push(k);
66             }
67             p[i] = a;
68         }
69     }
70     return;
71 }
72 int main(){
73     init();
74     int n;
75     while(~scanf("%d",&n)) while(n--){
76         char s1[5],s2[5];
77         int pre[5],after[5];
78         scanf("%s %s",s1,s2);
79         //
80         rep(i,0,4){
81             pre[i] = s1[i] - ‘0‘;
82             after[i] = s2[i] - ‘0‘;
83         }
84         //printf("%d %d\n",num(pre),num(after));
85         slove(num(pre),num(after));
86         //
87         if(!cur[num(after)]) printf("Impossible\n");
88         else printf("%d\n",val[num(after)]);
89     }
90     return 0;
91 }