Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 27132 | Accepted: 14861 |
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
Source
题目意思：
给出两个四位数的素数，要求对第一个素数进行一系列操作，使他变成第二个四位数的素数，每次操作只能改变一位数，且要求每次操作之后的数仍然是素数
问你最少的操作步骤数是多少？
分析：
先对四位数的素数打一个表
然后bfs
需要注意的是每次搜索入队之后，要进行回退操作
注意设置标记位（该数已经搜索过）
code：
#include<stdio.h> #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <math.h> #include <cstdlib> #include <queue> using namespace std; #define max_v 10100 int prime[max_v]; int vis[max_v]; struct node { int x,step; }; void init()//打N以内的素数表，prime[i]=0为素数。 { memset(prime,0,sizeof(prime)); prime[1]=1; for(int i=2; i<max_v; i++) { if(prime[i]==0) { for(int j=2; j*i<max_v; j++) prime[j*i]=1; } } } int bfs(int s,int e) { int num; memset(vis,0,sizeof(vis));//vis[N]用来标记是否查找过 queue<node> q; node p,next; p.x=s; p.step=0; vis[s]=1; q.push(p); while(!q.empty()) { p=q.front(); q.pop(); if(p.x==e) { return p.step; } int t[5]; t[1]=p.x/1000;//千 t[2]=p.x/100%10;//百 t[3]=p.x/10%10;//十 t[4]=p.x%10;//个 for(int i=1; i<=4; i++) { int temp=t[i];//这里特别注意，每次只能变换一位数，这里用来保存该变换位的数，变换下一位数时，该位数要恢复 for(int j=0; j<10; j++) { if(t[i]!=j) { t[i]=j;//换数 num=t[1]*1000+t[2]*100+t[3]*10+t[4]; } if(num>=1000&&num<=9999&&vis[num]==0&&prime[num]==0)//满足要求，四位数，没有用过，是素数 { next.x=num; next.step=p.step+1; q.push(next); vis[num]=1; } } t[i]=temp;//恢复 } } return -1; } int main() { //题意：给你两个4位数，要求你每次只能变换一位数字，并且变换前后的数字都要满足是素数；求最少变换次数； //首先打好素数表，再进行BFS int n,a,b,ans; init(); cin>>n; while(n--) { cin>>a>>b; ans=bfs(a,b); if(ans==-1) { cout<<"Impossible"<<endl; } else { cout<<ans<<endl; } } return 0; }