poj3126——bfs
POJ 1326 对数位的bfs
Prime Path
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12480 | Accepted: 7069 |
Description
![](http://poj.org/images/3126_1.jpg)
— 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
题意:求从一个四位数素数到另一个四位数素数所需的最短步数,每一步只能改变一位数字,且改变过程中的数必须是四位数
这是一道典型的数位搜索题,求最短路所以用bfs,难点在于模拟数的转化,另外注意边界控制
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; const int maxn=10000; int n,m; bool vis[maxn]; int ans; struct node { int num,step; }; int mypow(int n,int k) { int res=1; while(k--) res*=n; return res; } bool isprime(int n) //判断素数,由于数据不大,懒得打印素数表了 { for(int i=2;i*i<=n;i++){ if(n%i==0) return false; } return true; } int change(int n,int i,int j) //模拟数的转化,有很多技巧方法,一定要细心 { int f=n/mypow(10,i+1)*mypow(10,i+1);//保留前缀 int r=n%mypow(10,i);//保留后缀 int mid=j*mypow(10,i); //改变中间的数 return f+mid+r; } bool bfs() { memset(vis,0,sizeof(vis)); queue<node> q; q.push({n,0}); vis[n]=1; while(!q.empty()){ node now=q.front(); q.pop(); if(now.num==m){ ans=now.step; return true; } for(int i=0;i<4;i++){ //从第1位到第四位 for(int j=0;j<=9;j++){ //把所在位数字改为j int nextnum=change(now.num,i,j); if(nextnum<1000||nextnum>=maxn) continue;//边界控制 if(vis[nextnum]||!isprime(nextnum)) continue; vis[nextnum]=1; q.push({nextnum,now.step+1}); } } } return false; } int main() { int T;cin>>T; while(T--){ cin>>n>>m; if(bfs()) cout<<ans<<endl; else cout<<"Impossible"<<endl; } return 0; }
没有AC不了的题,只有不努力的ACMER!