【BFS】Prime Path-CFANZ编程社区

POJ 3126 Prime Path

Prime Path

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

6
7
0

题目大意

给你两个没有前导0的四位正整数 $n, m$ ，让你找出一条“路径”，使 $n$ 通过这条“路径”变成 $m$
“路径”包含若干个没有前导0的四位正整数，且每个数都是素数，这些素数都是由“路径”里的上一个数的某一位变化之后得来的，如 $8779 - > 8179$ ，8179就是由8779变化百位之后得来的，并且两个都是素数，再如题目所给的例子：
1033
1733
3733
3739
3779
8779
8179
然后题目所求的是最少变换次数，可以认为是“最短路径长度”

思路

BFS
对于每一个数，去变换它的每一位，每一位都随意变换（即在0~9之间任意取值）
那么每个数就可以变换出40个数，然后装在队列中，去重并BFS即可

代码如下

#include <iostream>
#include <queue>
#include <set>
using namespace std;

queue<pair<int, int> > q;
set<int> st;
pair<int, int> top;
int tmp;

//变换函数，把x从左往右数的第i+1个数字变成j（i从0到3）
int change(int x, int i, int j) {
    switch (i) {
        case 0:
            if (j == 0) return x;
            return j * 1000 + x % 1000;
        case 1:
            return (x / 1000) * 1000 + j * 100 + x % 100;
        case 2:
            return (x / 100) * 100 + j * 10 + x % 10;
        case 3:
            return (x / 10) * 10 + j;
        default:
            return -1;  //随便返回个什么数字
    }
}

//判断素数
bool is_prime(int n) {
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

// BFS
void bfs(int x, int y) {
    while (!q.empty()) q.pop();
    st.clear();
    q.push(make_pair(x, 0));
    st.insert(x);
    int i = 0, j = 0;
    while (!q.empty()) {
        top = q.front();
        q.pop();
        for (i = 0; i < 4; i++) {
            for (j = 0; j < 10; j++) {
                tmp = change(top.first, i, j);
                if (st.find(tmp) != st.end() || (!is_prime(tmp))) {
                    continue;
                }
                if (tmp == y) {
                    printf("%d\n", top.second + 1);
                    return;
                }
                q.push(make_pair(tmp, top.second + 1));
                st.insert(tmp);
            }
        }
    }
    printf("Impossible\n");
}

void solve() {
    int n, m, k;
    scanf("%d", &n);
    while (n--) {
        scanf("%d%d", &m, &k);
        if (m == k) {
            printf("0\n");
            continue;
        }
        bfs(m, k);
    }
}

int main(void) {
    solve();
    return 0;
}