| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 19095 | Accepted: 10711 |
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. 1033173337333739377987798179The 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
31033 81791373 80171033 1033Sample Output
670Source
Northwestern Europe 2006题目意思：
给两个是素数的四位数（千位始终不为0），每次改变其中一位，使得改变后的数依然是素数，计算从前一个数到后一个数最少一共需要改变几次。
解题思路：
打表1000~9999中所有的素数。
BFS：队列先将第一个数入队，然后分解出其各个位上的数，循环依次枚举个位~千位，可以是0~9中的任意一个数。
注意每次位数循环维护当前位不变，保证下次循环仅改变下一位上的数。
剪枝：①枚举到与当前位上的数相同的数，②处理过的数。
#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<queue>using namespace std;#define INF 0x3f3f3f3f#define MAXN 10000short int d[4];bool prime[MAXN],vis[MAXN];int cnt[MAXN];void init()//1000~9999中所有的素数{ memset(prime,false,sizeof(prime)); int i,j; for(i=1000; i<=10000; i++) { for(j=2; j<i; j++) if(i%j==0) { prime[i]=false; break; } if(j==i) prime[i]=true; }}int bfs(int a,int b){ queue <int> q; q.push(a); vis[a]=true; while(!q.empty()) { int s=q.front(); if(s==b) return cnt[s]; q.pop(); d[0]=s%10;//个位 d[1]=s%100/10;//十位 d[2]=s%1000/100;//百位 d[3]=s/1000;//千位 /*for(int j=3; j>=0; --j) cout<<d[j]<<" "; cout<<endl;*/ int temp,x; for(int j=0; j<4; ++j) //个位~千位 { x=d[j];//记录当前位置改变之前的数 for(int i=0; i<10; ++i) //0~9 { if(x==i) continue;//与当前位相同，剪枝 if(j==3&&i==0)continue;//千位不能为0 d[j]=i;//改变当前位的数 temp=d[3]*1000+d[2]*100+d[1]*10+d[0]; if(vis[temp]) continue;//已访问，剪枝 if(prime[temp])//是素数 { q.push(temp); vis[temp]=true; cnt[temp]=cnt[s]+1;//记录变化次数 } if(temp==b) return cnt[temp]; } d[j]=x;//维护当前位不变，保证下次循环仅改变下一位 } } return -1;}int main(){#ifdef ONLINE_JUDGE#else freopen("F:/cb/read.txt","r",stdin); //freopen("F:/cb/out.txt","w",stdout);#endif ios::sync_with_stdio(false); cin.tie(0); int t; cin>>t; init(); while(t--) { memset(vis,false,sizeof(vis)); memset(cnt,0,sizeof(cnt)); int a,b,ans=0; cin>>a>>b; ans=bfs(a,b); if(ans==-1) cout<<"Impossible"<<endl; else cout<<ans<<endl; } return 0;}/*31033 81791373 80171033 1033*/
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