POJ-3126 Prime Path

题意：给出两个四位素数a、b，问从a到b最少变几次（每次变1位并且只能为素数）。
思路：bfs。
View Code
 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <string>
 6 #include <queue>
 7 #include <algorithm>
 8 #include <iostream>
 9 using namespace std;
10 const int N=2000;
11 
12 typedef struct In{
13     int num;
14     int t;
15 }In;
16 In start;
17 queue<In>Q;
18 int a,b;
19 bool isprime[N*5],vis[N*5];
20 
21 int bfs(In c){
22     In now,next;
23     now=c;
24     Q.push(now);
25     while(!Q.empty()){
26         now=Q.front();
27         Q.pop();
28         if(now.num==b) return now.t;
29         for(int i=1;i<10;i++){
30             next.num=now.num/10*10+i;
31             if(!isprime[next.num]&&!vis[next.num]){
32                 vis[next.num]=true;
33                 next.t=now.t+1;
34                 Q.push(next);
35             }
36         }
37         for(int i=0;i<10;i++){
38             next.num=now.num/100*100+now.num%10+i*10;
39             if(!isprime[next.num]&&!vis[next.num]){
40                 vis[next.num]=true;
41                 next.t=now.t+1;
42                 Q.push(next);
43             }
44         }
45         for(int i=0;i<10;i++){
46             next.num=now.num/1000*1000+now.num%100+i*100;
47             if(!isprime[next.num]&&!vis[next.num]){
48                 vis[next.num]=true;
49                 next.t=now.t+1;
50                 Q.push(next);
51             }
52         }
53         for(int i=1;i<10;i++){
54             next.num=now.num%1000+i*1000;
55             if(!isprime[next.num]&&!vis[next.num]){
56                 vis[next.num]=true;
57                 next.t=now.t+1;
58                 Q.push(next);
59             }
60         }
61     }
62     return 0;
63 }
64 
65 int main(){
66     
67 //    freopen("data.in","r",stdin);
68 //    freopen("data.out","w",stdout);
69     
70     for(int i=2;i<=N*5;i++)
71         if(!isprime[i])
72             for(int j=2;i*j<=N*5;j++)
73                 isprime[i*j]=true;
74     int T;
75     scanf("%d",&T);
76     while(T--){
77         scanf("%d%d",&a,&b);
78         memset(vis,false,sizeof(vis));
79         vis[a]=true;
80         start.num=a; start.t=0;
81         while(!Q.empty()) Q.pop();
82         int ans=bfs(start);
83         printf("%d\n",ans);
84     }
85     return 0;
86 }
posted @ 2012-08-03 19:42 Hug_Sea 阅读(165) 评论(0) 收藏举报
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