来源:http://poj.org/problem?id=3126
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Prime Path
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. 1033The 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 |
题解:明显的广搜～～ 每次改变一位数，用hash判重解决之。
PS：这题出现了一个很蛋疼的错误，昨天样例都可以过， 但是提交WA。今天重写一遍-----发现把++exdir 写成exidir++囧..........
AC代码:
#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;
}
POj 3126 Prime Path,布布扣,bubuko.com
原文地址：http://blog.csdn.net/mummyding/article/details/38414167
