F – Prime PathTime Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 3126DescriptionThe 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.InputOne 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).OutputOne line for each case, either with a number stating the minimal cost or containing the word Impossible.Sample Input31033 81791373 80171033 1033Sample Output67
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#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>using namespace std;const int maxn=9999;const int inf=999999;int prim[maxn];int isprim[maxn];bool vis[maxn];int d[maxn];int num=0;bool pan(int a,int b){int ans=0;while(a){int k=a%10;int m=b%10;if(k!=m)ans++;a/=10;b/=10;}return ans==1;}int ans;void bfs(int s,int stop)//搜索{queue<int> q;memset(vis,0,sizeof(vis));q.push(s);vis[s]=true;d[s]=0;while(!q.empty()){int tmp=q.front();q.pop();// cout<<tmp<<endl;if(tmp==stop)return ;for(int i=0;i<num;i++){if(!vis[isprim[i]]&&pan(isprim[i],tmp)){vis[isprim[i]]=true;q.push(isprim[i]);d[isprim[i]]=d[tmp]+1;}}}}int main(){//freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout);int n,i,j,k,t;num=0;prim[0]=prim[1]=1;//筛选素数for(i=2;i<maxn;i++)if(!prim[i]){if(i>1000)isprim[num++]=i;for(j=i+i;j<maxn;j+=i)prim[j]=1;}int T;scanf("%d",&T);while(T–){int a,b;scanf("%d%d",&a,&b);d[b]=inf;bfs(a,b);if(d[b]!=inf)printf("%d\n",d[b]);elseprintf("Impossible\n");}return 0;}
，要知道，当你一直在担心错过了什么的时候，
