poj3126Prime Path （BFS+素数筛）

素数筛：需要一个数组进行标记

           最小的素数2，所有是2的倍数的数都是合数，对合数进行标记，然后找大于2的第一个非标记的数（肯定是素数），将其倍数进行标记，如此反复，若是找n以内的所有素数，只需要对[2,n^0.5]进行循环即可，因为n以内的所有数如果不是[2,n^0.5]的倍数，则一定是素数。复杂度：O(nloglogn);

for(int i=2;i*i<=n;i++){
if(a[i]!=-1)
for(int k=i*i;k<=n;k+=i)
a[i]=-1;
}

 

poj3126题目链接：http://poj.org/problem?id=3126

 

素数筛所占用的空间较大，毕竟是需要数组进行标记的，如果n太大就不可以做出（内存超出）。

若是让求[a,b]内的素数，a<b<=1e12,b-a<=1e6; 这个如果直接用素数筛是内存超出，可以先用一个数组存储[0,b^0.5]的素数，再用这些素数筛选[a,b],注意令一个数组表示[a,b]的素数，不能让下标直接表示这个数字（>=a&&<=b），而是开一个数组，c[i]表示a+i是否是素数。

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

现附上AC代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
using namespace std;
const int maxn=1e4+5;
int main(){
int T,num[4],temp;cin>>T;
int plu[maxn];
while(T--){
for(int i=0;i<=maxn;i++)
plu[i]=0;
for(int i=2;i*i<=maxn;i++){
for(int j=i*i;j<=maxn;j+=i){
plu[j]=-1;
}
}
int x,y;scanf("%d%d",&x,&y);
queue<int >q;
q.push(x);
plu[x]=1;//注意这一步的重要性，没有这一步的话，可能会导致 plu[temp]=plu[t]+1;当plu[t]==0时就执行了，
while(!q.empty()&&plu[y]==0){
int m=q.front(),t=q.front(); q.pop();

for(int i=3;i>=0;i--)
num[i]=m%10,m/=10;

for(int j=0;j<4;j++){
int mm=num[j],i;
if(j==0) i=1;
else i=0;
for(;i<=9;i++){
num[j]=i;
temp=num[0]*1000+num[1]*100+num[2]*10+num[3];
if(plu[temp]==0) {
plu[temp]=plu[t]+1;
q.push(temp);
}
}
num[j]=mm;
}
}
// printf("%d %d %d %d %d %d %d\n",plu[1033],plu[1733],plu[3733],plu[3739],plu[3779],plu[8779],plu[8179]);
if(plu[y]==-1) printf("Impossible\n");
else printf("%d\n",plu[y]-1);
}
return 0;
}

 

这道题是要求最小步骤到达目标，就是一个变形的最短路问题。对于最短路问题，肯定是首选BFS，而我们首先对x进行，个位十位百位千位的遍历，若新生成的数是素数（且之前没有访问过，毕竟是求最短路，若之前已经到过，则再到肯定不是最短，便剪枝），则将其放进队列。最后得出plu[y]基本就可以知道答案了。

 

回过头来一想，发现的确是很简单，但之前却wrong answer了很多次，自己都对自己无语了。