题目描述
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.
1033
1733
3733
3739
3779
8779
8179
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.
实际上就是给你两个数（都是素数），每一次可以变化第一个数的某一位（变化后也需要是素数），问最少变化几次可以变成第二个数。这里用一波bfs。
至于为什么用广搜，是因为他需要求最少的变化次数，所以搜索方法运用了广搜
输入
3
1033 8179
1373 8017
1033 1033输出
6
7
0贴一波代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <math.h>
#include <string>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#define maxn 1000000
const int MaxN=0x3f3f3f3f;
const int MinN=0xc0c0c00c;
typedef long long ll;
using namespace std;
bool isprim[100010];
void prim(){isprim[1]=false;for(int i=2;i<=100010;i++)if(isprim[i])for(int f=2*i;f<=100010;f=f+i)isprim[f]=false;
}
int num[10];
int x,y;
bool visited[100010];
int ans=0;
struct wazxy{int x;int steps;
}node,temp;queue<wazxy> q;void bfs(){node.x=x,node.steps=0;q.push(node);while(!q.empty()){temp=q.front();q.pop();if(temp.x==y){ans=temp.steps;return ;}else{int ii=0;while(temp.x){num[++ii]=temp.x%10;temp.x/=10;}// for(int i=1;i<=4;i++){
// cout<<num[i]<<" ";
// }
// cout<<endl;for(int i=1;i<=4;i++){ //分别用来分成改变个位十位百位千位的数字时for(int f=0;f<=9;f++){int now;if(i==1){if(f==num[1]) continue;now=num[4]*1000+num[3]*100+num[2]*10+f;//更改某位的数字，先将他拆开改变之后再合起来。if(isprim[now]&&visited[now]){ //我这个visited用来去重，防止变成某数又在变回去，而且一定要确定他是个素数node.steps=temp.steps+1;node.x=now;visited[now]=false; //标记q.push(node);}}if(i==2){if(f==num[2]) continue;now=num[4]*1000+num[3]*100+f*10+num[1];if(isprim[now]&&visited[now]){node.steps=temp.steps+1;node.x=now;visited[now]=false;q.push(node);}}if(i==3){if(f==num[3]) continue;now=num[4]*1000+f*100+num[2]*10+num[1];if(isprim[now]&&visited[now]){node.steps=temp.steps+1;node.x=now;visited[now]=false;q.push(node);}}if(i==4){if(f==0||f==num[4]) continue;//注意改变千位数字是不能为0now=f*1000+num[3]*100+num[2]*10+num[1];if(isprim[now]&&visited[now]){node.steps=temp.steps+1;node.x=now;visited[now]=false;q.push(node);}}}}}}
}int main()
{memset(isprim,true,sizeof(isprim));prim();//for(int i=0;i<=100;i++) if(isprim[i]) cout<<i<<" ";int t;cin>>t;for(int i=0;i<t;i++){memset(visited,true,sizeof(visited));cin>>x>>y;visited[x]=false;while(!q.empty()) q.pop(); //将队列清空！！！！！！！（之前一个题没有清空找了好久）ans=0;bfs();cout<<ans<<endl;}}
