CD0J/POJ 851/3126 方老师与素数/Prime Path BFS
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9982 | Accepted: 5724 |
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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 100001 const int inf=0x7fffffff; //无限大 const int MAXN = 10000; bool flag[MAXN]; int primes[MAXN], pi; struct point { int x; int y; }; void GetPrime_1() { int i, j; pi = 0; memset(flag, false, sizeof(flag)); for (i = 2; i < MAXN; i++) if (!flag[i]) { primes[i] = 1;//素数标识为1 for (j = i; j < MAXN; j += i) flag[j] = true; } } int vis[maxn]; int main() { GetPrime_1(); int t; cin>>t; while(t--) { memset(vis,0,sizeof(vis)); int n,m; cin>>n>>m; vis[n]=1; queue<point> q; q.push((point){n,0}); int flag1=0; while(!q.empty()) { point now=q.front(); if(now.x==m) { flag1=now.y; break; } point next; for(int i=0;i<=9;i++) { next.x=now.x/10; next.x*=10; next.x+=i; next.y=now.y+1; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push(next); } } for(int i=0;i<=9;i++) { int temp=now.x%10; next.x=now.x/100; next.x*=100; next.x+=i*10; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } for(int i=0;i<=9;i++) { int temp=now.x%100; next.x=now.x/1000; next.x*=1000; next.x+=i*100; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } for(int i=0;i<=9;i++) { int temp=now.x%1000; next.x=now.x/10000; next.x*=10000; next.x+=i*1000; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { // cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } if(flag1>0) break; q.pop(); } printf("%d\n",flag1); } return 0; }
