# Prime Path 分类： 搜索 POJ 2015-08-09 16:21 4人阅读 评论(0) 收藏 2022年11月23日
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Prime Path

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 14091 Accepted: 7959

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.

``1033173337333739377987798179``

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

Source

Northwestern Europe 2006

``#include <map>#include <list>#include <climits>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define eps 1e-9#define LL unsigned long long#define PI acos(-1.0)#define INF 0x3f3f3f3f#define CRR fclose(stdin)#define CWW fclose(stdout)#define RR freopen("input.txt","r",stdin)#define WW freopen("output.txt","w",stdout)const int Max = 10010;struct node{    int x;    int num;};int n,m;bool prime[Max];bool vis[Max];int bfs(){    memset(vis,false,sizeof(vis));    node a,b;    a.num=0;    a.x=n;    queue<node>Q;    vis[n]=true;    Q.push(a);    while(!Q.empty())    {        a=Q.front();        Q.pop();        if(a.x==m)        {            return a.num;        }        for(int i=0;i<10;i++)        {            b.x=a.x/10*10+i;            b.num=a.num+1;            if(!vis[b.x]&&!prime[b.x])            {                vis[b.x]=true;                Q.push(b);            }        }        for(int i=0;i<10;i++)        {            int s=a.x%10;            b.x=a.x/100*100+i*10+s;            b.num=a.num+1;            if(!vis[b.x]&&!prime[b.x])            {                vis[b.x]=true;                Q.push(b);            }        }        for(int i=0;i<10;i++)        {            int s=a.x%100;            b.x=a.x/1000*1000+i*100+s;            b.num=a.num+1;            if(!vis[b.x]&&!prime[b.x])            {                vis[b.x]=true;                Q.push(b);            }        }        for(int i=1;i<10;i++)        {            b.x=a.x%1000+i*1000;            b.num=a.num+1;            if(!vis[b.x]&&!prime[b.x])            {                vis[b.x]=true;                Q.push(b);            }        }    }    return 0;}int main(){    memset(prime,false,sizeof(prime));    prime=true;    prime=true;    m=sqrt(Max)+1;    for(int i=2;i<m;i++)    {        if(!prime[i])        {            for(int j=i*i;j<=Max;j+=i)            {                prime[j]=true;            }        }    }    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&m);        printf("%d\n",bfs());    }    return 0;}`` 微信扫一扫 支付宝扫一扫 python开发_常用的python模块及安装方法 Educational Codeforces Round 11 C. Hard Process 二分
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