POJ 3126 Prime Path(BFS)

最新推荐文章于 2021-10-11 16:57:25 发布

ToRe.

最新推荐文章于 2021-10-11 16:57:25 发布

阅读量157

点赞数

分类专栏： # 搜索文章标签： bfs

本文链接：https://blog.csdn.net/j2_o2/article/details/78946071

版权

搜索专栏收录该内容

24 篇文章

订阅专栏

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23967		Accepted: 13242

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

Sample Output

6
7
0

思路：暴力广搜每次能改变数的可能，为了使代码精简些，牺牲了一点运行速度（将条件判断移到了每次出队而不是入队）

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<queue>
using namespace std;
#define mod 1000000007
#define ll long long

int pri[10005],book[10005];//pri记录素数，book记录该数是否出现过

struct node
{
    int id,cnt;
}st,ne;

//筛素数
void prime_number(){
    int i,j;
    pri[0]=1;
    pri[1]=1;
    for(i=2;i<101;i++)
        if(!pri[i])
            for(j=i+i;j<10000;j+=i)
                pri[j]=1;
}

int main(){
    prime_number();
    int T;
    scanf("%d",&T);
    while(T--){
        memset(book,0,sizeof(book));
        int m,i,s,TS;
        scanf("%d%d",&st.id,&m);
        st.cnt=0;
        book[st.id]=1;
        queue<node> q;
        q.push(st);

        while(!q.empty()){
            st=q.front();
            if(st.id==m)break;
            q.pop();
            ne.cnt=st.cnt+1;
            //个位
            for(i=0;i<10;i++){
                ne.id=st.id/10*10+i;
                if(pri[ne.id]||book[ne.id])continue;
                book[ne.id]=1;
                q.push(ne);
            }

            //十位
            for(i=0;i<10;i++){
                TS=st.id%10;
                ne.id=st.id/100*100+i*10+TS;
                if(pri[ne.id]||book[ne.id])continue;
                book[ne.id]=1;
                q.push(ne);
            }

            //百位
            for(i=0;i<10;i++){
                TS=st.id%100;
                ne.id=st.id/1000*1000+i*100+TS;
                if(pri[ne.id]||book[ne.id])continue;
                book[ne.id]=1;
                q.push(ne);
            }

            //千位
            for(i=1;i<10;i++){
                TS=st.id%1000;
                ne.id=i*1000+TS;
                if(pri[ne.id]||book[ne.id])continue;
                book[ne.id]=1;
                q.push(ne);
            }
        }
        if(st.id==m)printf("%d\n",st.cnt);
        else printf("Impossible\n");
    }
    return 0;
}