POJ 3216 Prime Path（数字BFS）

Prime Path

 

大意：给你两个数，求从第一个数经过几次变换到第二个数。变换要求：1.中间数必须是素数   2.每次只能变一个数字。

思路：对每一位数字进行bfs

 

 

#include <map>
#include <stack>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <limits.h>
#include <algorithm>
#define LL long long
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
#define eps 1e-9
#define INF 1 << 30
using namespace std;

int a, b, Ans;
bool flag[10000];
int vis[10000] = {0};
void prime()
{
    int k, j;
    for(int i = 1001; i < 10000; i+=2)
    {
        k = sqrt(i*1.0);
        for(j = 2; j <= k; j++)
            if(i % j == 0)
                break;
        if(j >= k + 1)
            flag[i] = 1;
    }
}

void BFS()
{
    int t;
    int i;
    int p1, p2, p3;
    queue<int> q;
    q.push(a);
    vis[a] = true;
    p1 = 0;
    p2 = p3 = 1;    
    while (!q.empty())
    {
        t = q.front();
        q.pop();
        p1++;
        if (t == b)
            return;
        int next;
        for (i = 1; i <= 9; i++)
        {
            next = t%1000;
            next += i*1000;
            if (flag[next] && !vis[next])
            {
                vis[next] = true;
                q.push(next);
                p2++;
            }
        }
        for (i = 0; i <= 9; i++)
        {
            int temp;
            temp = t/100%10;
            next = t-temp*100;
            next += i*100;
            if (flag[next] && !vis[next])
            {
                vis[next] = true;
                q.push(next);
                p2++;
            }
        }
        for (i = 0; i <= 9; i++)
        {
            int temp;
            temp = t/10%10;
            next = t-temp*10;
            next += i*10;
            if (flag[next] && !vis[next])
            {
                vis[next] = true;
                q.push(next);
                p2++;
            }
        }
        for (i = 0; i <= 9; i++)
        {
            int temp;
            temp = t%10;
            next = t-temp;
            next += i;
            if (flag[next] && !vis[next])
            {
                vis[next] = true;
                q.push(next);
                p2++;
            }
        }
        if (p1 == p3)
        {
            Ans++;
            p3 = p2;
        }
    }
}

void Solve()
{
    int n;
    cin >> n;
    while(n--)
    {
        cin >> a >> b;
        memset(vis, 0, sizeof(vis));
        Ans = 0;
        BFS();
        cout << Ans << endl;
    }
}

int main(void)
{
    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    prime();
    Solve();

    return 0;
}