poj 3126 Prime Path(bfs)

标签：poj 3126 prime path   bfs   

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11932   Accepted: 6772 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 Source Northwestern Europe 2006

题意：

给你n,m,都是四位数，求最少步数使n转化为m.

要求：每次只能改一位数，而且改完后的数必须是素数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

bool vis[10010],is_prime[10010];
struct node
{
    int x;
    int num;
};
queue<node>que;
int n,m;
int flag;

void Prime()
{
    memset(is_prime,0,sizeof is_prime);
    is_prime[0]=is_prime[1]=1;
    for(int i=2; i<=10000; i++)
    {
        if(!is_prime[i])
        {
            for(int j=i+i; j<=10000; j+=i)
                is_prime[j]=1;
        }
    }
}

int pow_1(int n,int k)
{
    int sum=1;
    for(int i=0;i<k;i++)
        sum*=n;
    return sum;
}

void bfs()
{
    memset(vis,0,sizeof vis);
    while(que.size())
        que.pop();
    node a,t;
    vis[n]=1;
    a.x=n;
    a.num=0;
    que.push(a);
    while(que.size())
    {
        a=que.front();
        que.pop();
        if(a.x==m)
        {
            flag=0;
            cout<<a.num<<endl;
            return;
        }
        int c[5];
        int l=0;
        int num=a.x;
        while(num)
        {
            c[l++]=num%10;
            num/=10;
        }
        for(int i=0; i<l; i++)
        {
            for(int j=0; j<10; j++)
            {
                int newn=0;
                for(int k=0; k<l; k++)
                {
                    if(k==i)//要换的数位
                        newn+=pow_1(10,k)*j;
                    else               ///其他不变数位
                        newn+=pow_1(10,k)*c[k];
                }
                if(newn>=1000&&newn<10000&&!is_prime[newn]&&!vis[newn])
                {
                    t.x=newn,t.num=a.num+1;
                    que.push(t);
                    vis[newn]=1;
                }
            }
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    Prime();
    while(cin>>T)
    {
        while(T--)
        {
            cin>>n>>m;
            flag=1;
            bfs();
            if(flag)
                cout<<"Impossible\n";
        }
    }
    return 0;
}

poj 3126 Prime Path(bfs)

标签：poj 3126 prime path   bfs   

原文地址：http://blog.csdn.net/acm_baihuzi/article/details/42376779