POj 3126 Prime Path

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來源:http://poj.org/problem?id=3126

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11384   Accepted: 6453 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 31033 81791373 80171033 1033 Sample Output 670 Source Northwestern Europe 2006

題目大意:  從素數四位元a變換到四位元b，每次只能改變一位元字，且每次改變後的數也必須為素數～～ 四位元中第一位不能為0，求最少變換次數，如果無法變換，則輸出"Impossible".

題解:明顯的廣搜～～ 每次改變一位元，用hash判重解決之。

PS：這題出現了一個很蛋疼的錯誤，昨天範例都可以過， 但是提交WA。今天重寫一遍-----發現把++exdir 寫成exidir++囧..........

 AC代碼:

#include<iostream>#include<cmath>#include<cstring>using namespace std;const int Max=2500000,Maxh=10005;bool visit[Maxh]={0},Isprime[Maxh]={0};int  star,dest;bool prime(int k){    for(int i=2;i<=(int)sqrt((float)k);i++)    if(k%i==0)    return false;    return true;}struct Node{    int dig[4],num,step;}map[Max]={0},temp;void calc(Node &x){x.dig[0]=x.num/1000;x.dig[1]=(x.num/100)%10;x.dig[2]=(x.num/10)%10;x.dig[3]=x.num%10;}void calcsum(Node &x){    x.num=0;    for(int i=0;i<4;i++)    x.num+=x.dig[i]*(int)pow((float)10,3-i);}int main(){    int t;    cin>>t;    for(int i=1000;i<10000;i++)    if(prime(i))    Isprime[i]=1;    while(t--){    memset(visit,0,sizeof(visit));    int flag=1;    cin>>star>>dest;    if(star==dest){    cout<<0<<endl;    continue;    }    map[0].num=star;  calc(map[0]);  map[0].step=0;    int nodedir=0,exdir=0;    while(nodedir<=exdir&&exdir<Max&&flag){    for(int i=0;i<4&&flag;i++){    for(int k=0;k<10;k++){    if(i||k){    if(i==3&&k%2==0)    continue;    temp=map[nodedir];    temp.dig[i]=k;    calcsum(temp);    temp.step++;    if(temp.num==dest){    cout<<temp.step<<endl;    flag=0;    break;    }    if(Isprime[temp.num]&&!visit[temp.num]){    visit[temp.num]=1;    map[++exdir]=temp;    }    }    }    }    nodedir++;    }    if(flag)    cout<<"Impossible"<<endl;    }return 0;}