POJ 3126 Prime Path
Description
Now, the minister of finance, who had been eavesdropping, intervened.
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased. Input One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros). Output One line for each case, either with a number stating the minimal cost or containing the word Impossible. Sample Input 3 1033 8179 1373 8017 1033 1033 Sample Output 6 7 0 大致题意： 给定两个四位素数a b，要求把a变换到b 变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。 求从a到b最少需要的变换次数。无法变换则输出Impossible 注意：双向广搜是在一个队列中实现的，只不过是交替进行罢了！ |
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. 1 #include<iostream>
2 using namespace std;
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<cstdlib>
7 #include<queue>
8 #define N 10000
9 struct prime{
10 int c[4];
11 int flag;
12 };
13 int dis[N];
14 int visit[N];
15 queue<prime>que;
16 int js(int *m)
17 {
18 return (m[0]*1000+m[1]*100+m[2]*10+m[3]*1);
19 }
20 bool is_prime(int l)
21 {
22 bool flag=true;
23 for(int i=2;i<=sqrt(l);++i)
24 {
25 if(l%i==0)
26 {
27 flag=false;
28 break;
29 }
30 }
31 return flag;
32 }
33 int bfs()
34 {
35 while(!que.empty())
36 {
37 prime x=que.front();
38 que.pop();
39 int now=js(x.c);
40 for(int i=1;i<=9;++i)
41 {
42 prime nx=x;
43 nx.c[0]=i;
44 int shu=js(nx.c);
45 if(!visit[shu]&&is_prime(shu))
46 {
47 visit[shu]=x.flag;
48 nx.flag=x.flag;
49 que.push(nx);
50 dis[shu]=dis[now]+1;
51 }
52 else if(visit[shu]&&visit[shu]!=x.flag)
53 {
54 return dis[now]+dis[shu]+1;
55 }
56 }
57 for(int j=1;j<=3;++j)
58 {
59 for(int i=0;i<=9;++i)
60 {
61 prime nx=x;
62 nx.c[j]=i;
63 int shu=js(nx.c);
64 if(!visit[shu]&&is_prime(shu))
65 {
66 visit[shu]=x.flag;
67 nx.flag=x.flag;
68 que.push(nx);
69 dis[shu]=dis[now]+1;
70 }
71 else if(visit[shu]&&visit[shu]!=x.flag)
72 {
73 return dis[now]+dis[shu]+1;
74 }
75 }
76
77 }
78 }
79 return -1;
80 }
81 int main()
82 {
83 int tex;
84 scanf("%d",&tex);
85 char a[10];
86 while(tex--)
87 {
88 while(!que.empty()) que.pop();
89 memset(dis,0,sizeof(dis));
90 memset(visit,0,sizeof(visit));
91 scanf("%s",a);
92 que.push(prime{a[0]-‘0‘,a[1]-‘0‘,a[2]-‘0‘,a[3]-‘0‘,1});
93 int p=(a[0]-‘0‘)*1000+(a[1]-‘0‘)*100+(a[2]-‘0‘)*10+(a[3]-‘0‘);
94 visit[p]=1;dis[p]=0;
95 int q=p;
96 scanf("%s",a);
97 que.push(prime{a[0]-‘0‘,a[1]-‘0‘,a[2]-‘0‘,a[3]-‘0‘,2});
98 p=(a[0]-‘0‘)*1000+(a[1]-‘0‘)*100+(a[2]-‘0‘)*10+(a[3]-‘0‘);
99 visit[p]=2;dis[p]=0;
100 if(q==p)
101 {
102 printf("0\n");continue;
103 }
104 int temp=bfs();
105 if(temp==-1) printf("Impossible\n");
106 else printf("%d\n",temp);
107 }
108 return 0;
109 }