POJ-3126 Prime Path

时间：2018-07-28 18:19:26 阅读：142 评论：0 收藏：0 [点我收藏+]

标签：use note rom iostream str heap please nbsp 16px

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

Sample Output

6
7
0

大致题意：多组样例，给定两个四位质数，作为起点和终点，要求每次改变一位数字，改变后的数字仍为质数，输出最少的改变步数

可以先把所有的质数先筛选出来，之后bfs，每次改变一位数字并判断其是否为质数，直至到达终点，代码如下：

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 void read(int &x)
 8 {
 9     int f=0;x=0;char s=getchar();
10     while(s>‘9‘||s<‘0‘){if(s==‘-‘)f=1;s=getchar();}
11     while(s>=‘0‘&&s<=‘9‘){x=(x<<1)+(x<<3)+(s^48);s=getchar();}
12     x=f?-x:x;
13 }
14 int t,x,y,cnt=0,prime[10005],a,b,c,d,ans;
15 bool num[10005],vis[10005];
16 struct node{
17     int k,cost;
18 };
19 int bfs()
20 {
21     queue<node>q;
22     node now,next;
23     now.k=x,now.cost=0,vis[now.k]=1;
24     q.push(now);
25     while(!q.empty())
26     {
27         now=q.front();
28         q.pop();
29         if(now.k==y) return now.cost;
30         a=now.k/1000,b=now.k/100%10,c=now.k/10%10,d=now.k%10;
31         for(int i=1;i<=9;++i)
32         {
33             if(i==a) continue;
34             next.k=i*1000+b*100+c*10+d;
35             if(vis[next.k]||num[next.k]==0) continue;
36             next.cost=now.cost+1,vis[next.k]=1;
37             q.push(next);
38         }
39         for(int i=0;i<=9;++i)
40         {
41             if(i==b) continue;
42             next.k=a*1000+i*100+c*10+d;
43             if(vis[next.k]||num[next.k]==0) continue;
44             next.cost=now.cost+1,vis[next.k]=1;
45             q.push(next);
46         }
47         for(int i=0;i<=9;++i)
48         {
49             if(i==c) continue;
50             next.k=a*1000+b*100+i*10+d;
51             if(vis[next.k]||num[next.k]==0) continue;
52             next.cost=now.cost+1,vis[next.k]=1;
53             q.push(next);
54         }
55         for(int i=0;i<=9;++i)
56         {
57             if(i==d) continue;
58             next.k=a*1000+b*100+c*10+i;
59             if(vis[next.k]||num[next.k]==0) continue;
60             next.cost=now.cost+1,vis[next.k]=1;
61             q.push(next);
62         }
63     }
64     return 0;
65 }
66 int main()
67 {
68     memset(num,1,sizeof(num));
69     num[0]=num[1]=0;
70     for(int i=2;i<=10000;++i)
71     {
72         if(num[i]) prime[++cnt]=i;
73         for(int j=1;j<=cnt&&prime[j]*i<=10000;++j)
74         {
75             num[prime[j]*i]=0;
76             if(!(i%prime[j])) break;
77         }
78     }
79     read(t);
80     while(t--)
81     {
82         memset(vis,0,sizeof(vis));
83         read(x),read(y);
84         if(x==y)
85         {
86             printf("0\n");
87             continue;
88         }
89         ans=bfs();
90         if(ans) printf("%d\n",ans);
91         else printf("Impossible\n");
92     }
93     return 0;
94 }

POJ-3126 Prime Path

标签：use note rom iostream str heap please nbsp 16px

原文地址：https://www.cnblogs.com/n0-b0dy/p/9382706.html

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