POJ 3126 Prime Path （bfs+欧拉线性素数筛）

时间：2016-11-06 03:01:41 阅读：198 评论：0 收藏：0 [点我收藏+]

标签：++ nec which security pos using stat path stp

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

Sample Output

6
7
0

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 }

POJ 3126 Prime Path （bfs+欧拉线性素数筛）

标签：++ nec which security pos using stat path stp

原文地址：http://www.cnblogs.com/agenthtb/p/6034484.html

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