题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1973
Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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.
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).
the minimal cost or containing the word Impossible.
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
struct node{
int cur, step;
}now, Next;
int vis[], star, finish, prime[] = { , , };
void init(){
for (int i = ; i < ; i++){
if (!prime[i]){
for (int j = ; i*j < ; j++)
prime[i*j] = ;
}
}
}
int bfs(){
queue<node> Q;
vis[star] = ;
now.cur = star, now.step = ;
Q.push(now);
while (!Q.empty()){
int i, j;
char num[];
now = Q.front();
Q.pop();
if (now.cur == finish) return now.step;
for (i = ; i < ; i++){
sprintf(num, "%d", now.cur);
for (j = ; j < ; j++){
if (j == && i == )
continue;
if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j * + (num[] - '') * + (num[] - '') * + (num[] - '');
else if (i == )
Next.cur = j + (num[] - '') * + (num[] - '') * + (num[] - '') * ;
if (!prime[Next.cur] && !vis[Next.cur])
{
Next.step = now.step + ;
vis[Next.cur] = ;
Q.push(Next);
}
}
}
}
return -;
}
int main(){
int t, ans;
cin >> t;
init();
while (t--){
cin >> star >> finish;
memset(vis, , sizeof(vis));
ans = bfs();
if (ans == -) cout << "Impossible\n";
else cout << ans << endl;
}
return ;
}
其实这道题学校OJ(Swust OJ)也有但是坑爹的后台数据变成a+b的后台数据了，Orz~~~(无爱了)