POJ

题目链接：https://vjudge.net/problem/POJ-3126

从一个素数到另一个素数，每一步只能改变一个数，途径的数也必须是素数，求最小步数。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>

using namespace std;
typedef long long ll;
const int maxn = 2e6 + 5;
const int INF = 0x3f3f3f3f;

bool isPrime[10005];
int vis[10005];
int prime[10005];

void Prime(int R)
{
    memset(isPrime, 0, sizeof(isPrime));
    memset(prime, 0, sizeof(prime));
    for (int i = 2; i <= R; i++) {
        if (!isPrime[i]) {
            prime[++prime[0]] = i;
        }
        for (int j = 1; j <= prime[0] && i * prime[j] <= R; j++) {
            isPrime[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}

struct node
{
    int num;
    int step;
};

int st, ed;

int bfs(int st)
{
    queue<node>qe;
    node start, next;
    start.num = st;
    start.step = 0;
    memset(vis, 0, sizeof(vis));
    qe.push(start);
    while (!qe.empty()) {
        node temp = qe.front();
        qe.pop();
        if (temp.num == ed) {
            return temp.step;
        }
        vis[temp.num] = 1;
        for (int k = 0; k < 4; k++) {
            int head = 0;
            int x = (int)pow(10.0, k);
            head = temp.num - (temp.num / x % 10) * x;
            for (int i = 0; i <= 9; i++) {
                if (x == 1000 && i == 0) continue;
                next.num = head + i * x;
                if (next.num > 9999) break;
                if (!isPrime[next.num] && !vis[next.num]) {
                    next.step = temp.step + 1;
                    qe.push(next);
                    vis[next.num] = 1;
                }
            }
        }
    }
    return -1;
}

int main()
{
    Prime(10001);
    int t;
    cin >> t;
    while (t--)
    {
        cin >> st >> ed;
        int ans = bfs(st);
        cout << ans << endl;
    }
    return 0;
}