shuoj1019-Prime Path--bfs

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.

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.

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

Sample Output

题意：：从n换到m，每次换n中的一位，每一次换后都为质数，求最少换的次数。

题解：：明显的本题需要使搜索，dfs或bfs。dfs也想了很久没有解答出来。bfs本身就可以求最短路径，适合本题求解。

#include<bits/stdc++.h> using namespace std; #define XINF INT_MAX #define INF 0x3FFFFFFF #define MP(X,Y) make_pair(X,Y) #define PB(X) push_back(X) #define REP(X,N) for(int X=0;X<N;X++) #define REP2(X,L,R) for(int X=L;X<=R;X++) #define DEP(X,R,L) for(int X=R;X>=L;X--) #define CLR(A,X) memset(A,X,sizeof(A)) #define IT iterator #define LOCAL typedef long long ll; typedef pair<int,int> PII; typedef vector<PII> VII; typedef vector<int> VI; //const int MAXN = 10010; //判断素数 bool sushu(int x){ for(int i = 2;i*i<=x;i++) if(x%i==0) return 0; return 1; } int vis[10000];//判断是否加入过队列 int cnt[10000];//cnt[i]记录从n到i的操作次数。。。。为本题关键 int main(){ /* #ifdef LOCAL freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif */ int T; cin>>T; while(T--){ int n,m; cin>>n>>m; memset(vis ,0 ,sizeof(vis)); memset(cnt ,0 ,sizeof(cnt)); queue<int> Q; Q.push(n); while(!Q.empty()){ int temp = Q.front();Q.pop(); vis[temp] = 1; if(temp == m)break;//当找到temp == m是即可退出循环 int i = 0; int x = temp; int a[5];//记录temp的各个位数大小 while(temp){ a[i++] = temp%10; temp/=10; } REP(i,10){ REP(j,4){ if(j==3&&i==0)continue;//千位为零时不进行运算 if(j==0) temp = i+a[1]*10+a[2]*100+a[3]*1000;//个位换 if(j==1)temp = a[0]+i*10+a[2]*100+a[3]*1000;//十位 if(j==2)temp = a[0]+a[1]*10+i*100+a[3]*1000;//百位 if(j==3)temp = a[0]+a[1]*10+a[2]*100+i*1000;//千位 if(vis[temp] == 0&&sushu(temp) == 1) { vis[temp] = 1; cnt[temp] = cnt[x]+1;//从前一个数变到现在的数操作次数加一 Q.push(temp); } } } } if(cnt[m]>0||n == m) cout<<cnt[m]<<endl;//输出从n到m的操作次数 else cout<<"Impossible"<<endl; } //printf("Time used = %.2lf\n",(double)clock()/CLOCKS_PER_SEC); }