Description
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.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime.
You will just have to paste four new digits over the four old ones on
your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to
8179 by a path of prime numbers where only one digit is changed from one
prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don‘t know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest
going on... Help the prime minister to find the cheapest prime path
between any two given four-digit primes! The first digit must be
nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
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 70 题意给你两个4位素数a b，让你将a每次改变一位数字，改变后的4位数还必须是素数，最少几步能变到b，输出步数，不能变到输出Impossible。这题第一发T了...一看就是果然T，忘了加vis2这个数组标记那些数组被访问过了...不加的话因为队列及时pop了，会出现在两个素数之间来回跳的情况！小插曲就是欧拉线性素数筛，自己还没背会2333333。代码如下：
1 #include <iostream>
2 #include <cstdio>
3 #include <queue>
4 #include <algorithm>
5 #include <cstring>
6 using namespace std;
7 #define inf 0x3f3f3f3f
8 int prime[11000],a,b,ans;
9 bool num[11000],vis[11000],vis2[11000];
10 bool flag;
11 int dig[4];
12 struct node
13 {
14 int x,stp;
15 };
16 void split (int x)//将x分成各个位
17 {
18 for (int i=0;i<4;++i)
19 {
20 dig[i]=x%10;
21 x/=10;
22 }
23 }
24 int getnum (int a,int b,int c,int d)//将4个数字组成一个四位数
25 {
26 return a+b*10+c*100+d*1000;
27 }
28 void getprime()//欧拉线性素数筛
29 {
30 memset(num,false,sizeof num);
31 memset(vis,false,sizeof vis);
32 memset(prime,0,sizeof prime);
33 int cnt=0;
34 for (int i=2;i<11000;++i)
35 {
36 if (!vis[i])
37 prime[cnt++]=i,num[i]=1;
38 for (int j=0;j<cnt&&i*prime[j]<11000;++j)
39 {
40 vis[i*prime[j]]=1;
41 if (i%prime[j]==0)
42 break;
43 }
44 }
45 }
46 void bfs(node now)
47 {
48 queue<node>q;
49 q.push(now);
50 vis2[now.x]=true;
51 if (now.x==b)
52 {
53 flag=true;
54 ans=now.stp;
55 return ;
56 }
57 while (!q.empty())
58 {
59 node frt=q.front();
60 q.pop();
61 if (frt.x==b)
62 {
63 flag=true;
64 ans=frt.stp;
65 return ;
66 }
67 split(frt.x);
68 for (int i=0;i<=9;++i)//改个位
69 {
70 int temp=getnum(i,dig[1],dig[2],dig[3]);
71 if (temp==frt.x)
72 continue;
73 if (num[temp]&&!vis2[temp])
74 {
75 node tp;
76 tp.x=temp;
77 tp.stp=frt.stp+1;
78 vis2[temp]=true;
79 q.push(tp);
80 }
81 }
82 for (int i=0;i<=9;++i)//改十位
83 {
84 int temp=getnum(dig[0],i,dig[2],dig[3]);
85 if (temp==frt.x)
86 continue;
87 if (num[temp]&&!vis2[temp])
88 {
89 node tp;
90 tp.x=temp;
91 tp.stp=frt.stp+1;
92 vis2[temp]=true;
93 q.push(tp);
94 }
95 }
96 for (int i=0;i<=9;++i)//改百位
97 {
98 int temp=getnum(dig[0],dig[1],i,dig[3]);
99 if (temp==frt.x)
100 continue;
101 if (num[temp]&&!vis2[temp])
102 {
103 node tp;
104 tp.x=temp;
105 tp.stp=frt.stp+1;
106 vis2[temp]=true;
107 q.push(tp);
108 }
109 }
110 for (int i=1;i<=9;++i)//改千位，注意是4位数，所以千位不为0，从一开始
111 {
112 int temp=getnum(dig[0],dig[1],dig[2],i);
113 if (temp==frt.x)
114 continue;
115 if (num[temp]&&!vis2[temp])
116 {
117 node tp;
118 tp.x=temp;
119 tp.stp=frt.stp+1;
120 vis2[temp]=true;
121 q.push(tp);
122 }
123 }
124 }
125 return ;
126 }
127 int main()
128 {
129 getprime();
130 //freopen("de.txt","r",stdin);
131 int t;
132 scanf("%d",&t);
133 while (t--)
134 {
135 scanf("%d%d",&a,&b);
136 memset(vis2,false,sizeof vis2);
137 ans=inf;
138 node now;
139 now.x=a,now.stp=0;
140 bfs(now);
141 if (flag)
142 printf("%d\n",ans);
143 else
144 printf("Impossible\n");
145 }
146 return 0;
147 }