EOJ1646 bfs
Time Limit:1000MSMemory Limit:65536KB
Total Submit:68Accepted:40
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
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
方品 10121540117
题目：EOJ1646
题目分析：
最短路径问题，想到用队列实现bfs求解。可构造一个结构体存储四位数的每一位。先求出四位数的质数，这样可在o（1）时间内判断一个数是否为素数。将第一个数放入队列，每次取出队首元素，依次改变其各位，若为质数，则进队。注意这里可以剪枝，若新的质数在之前已经遍历过了则不再进队。可通过一个hash数组标记该数是否被访问过，若对手元素为所求元素则返回路径长度，若队空跳出循环则说明不存在答案。
AC代码：
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
bool prime[10005];
bool used[10005];
struct Num{
int a[4],step;
Num(){memset(a,0,sizeof(a));step=0;}
int val(){
return a[0]*1000+a[1]*100+a[2]*10+a[3];
}
};
void setprime(){
int i,j;
for(i=0;i<10005;++i) prime[i]=i;
i=2;
while(i<101){
if(prime[i])
for(j=2;i*j<10005;++j)
prime[i*j]=false;
++i;
}
}
int bfs(string a,string b){
queue
Num now,cmp;
for(int i=0;i<4;++i){
now.a[i]=a[i]-'0';
cmp.a[i]=b[i]-'0';
}
q.push(now);
used[now.val()]=true;
while(!q.empty()){
now=q.front();
q.pop();
for(int i=0;i<4;++i){
Num next=now;
for(int j=0;j<=9;++j){
if(i==0 && j==0)continue;
next.a[i]=j;
next.step=now.step+1;
if(next.val()==cmp.val()){
return next.step;
}
if( !used[next.val()]&& prime[next.val()] ){
q.push(next);
used[next.val()]=true;
}
}
}
}
return -1;
}
int main()
{
int t;
setprime();
cin>>t;
while(t--){
memset(used,false,sizeof(used));
string a,b;
cin>>a>>b;
if(a==b)
cout<<"0"<
else{
int ans=bfs(a,b);
if(ans!=-1)
cout<
else
cout<<"Impossible"<
}
}
return 0;
}