D-POJ-3126 Prime Path
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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 1033Sample Output
6
7
0
BFS,将素数打表可以缩短运行时间。
#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<cstring>using namespace std;#define maxn 10005bool vis[maxn]; //标记是否访问bool IsPrime[maxn]; //标记素数void Prime(){ //素数筛法 IsPrime[0]=IsPrime[1]=false;IsPrime[2]=true; for(int i=3;i<maxn;i++){ if(i%2) IsPrime[i]=true; else IsPrime[i]=false; } int m=sqrt(maxn*1.0); for(int i=3;i<m;i++){ if(IsPrime[i]){ for(int j=i;j<maxn;j+=i) IsPrime[j]=false; } }}struct Node{ int p[4]; int step;};void BFS(int a,int b){ memset(vis, false, sizeof(vis)); Node now,next; int x; queue<Node>Q; now.p[0]=a/1000;now.p[1]=a%1000/100;now.p[2]=a%100/10;now.p[3]=a%10; now.step=0; Q.push(now); while(!Q.empty()){ now=Q.front(); Q.pop(); if(now.p[0]*1000+now.p[1]*100+now.p[2]*10+now.p[3]==b){ printf("%d\n",now.step); return; } for(int i=0;i<=3;i++){ for(int j=0;j<=9;j++){ if(i==0&&j==0) continue; if(now.p[i]==j) continue; next.p[0]=now.p[0]; next.p[1]=now.p[1]; next.p[2]=now.p[2]; next.p[3]=now.p[3]; next.p[i]=j; x=next.p[0]*1000+next.p[1]*100+next.p[2]*10+next.p[3]; if(!vis[x]&&IsPrime[x]){ next.step=now.step+1; vis[x]=true; Q.push(next); } } } } printf("Impossible\n");}int main(){ Prime(); int n,a,b; scanf("%d",&n); while(n--){ scanf("%d%d",&a,&b); BFS(a,b); } return 0;}
开始不管怎么运行都是输出Impossible,后来发现Prim函数没运行,蠢哭~
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