poj 3126 Prime Path（bfs）

题意：给两个四位素数，从第一个素数变换到第二个素数最少需要变换多少次，每次只变换一个位置上的数字，第一位数字不能为0

简单bfs

#include <cstdio>#include <cstring>#include <queue>using std::queue;struct node{    int val;    int step;};bool prime[10000];bool book[10000];int s,e,res;int num[4];void getPrime(){    memset(prime,0,sizeof(prime));    prime[0] = prime[1] = true;    for(int i = 2; i < 10000; ++i)    {        if(!prime[i])            for(int j = 2*i; j < 10000; j += i)                prime[j] = true;//false表示是素数，true表示不是素数    }}bool isPrime(int num){    return !prime[num];}void BFS(){    memset(book,0,sizeof(book));    queue<node> que;    node head;    head.val = s;    head.step = 0;    book[s] = true;    que.push(head);    int temp;    while(!que.empty())    {        node cur = que.front();        que.pop();        if(cur.val == e)        {            res = cur.step;            return;        }        num[0] = cur.val/1000;        num[1] = (cur.val/100)%10;        num[2] = (cur.val/10)%10;        num[3] = cur.val%10;        for(int i = 0; i < 4; ++i)        {            temp = num[i];            for(int j = 0; j < 10; ++j)            {                if(i == 0 && j == 0) continue;                if(temp != j)                {                    num[i] = j;                    int n = num[0]*1000+num[1]*100+num[2]*10+num[3];                    if(!book[n] && isPrime(n))                    {                        node atp;                        atp.val = n;                        atp.step = cur.step + 1;                        book[n] = true;                        que.push(atp);                    }                }            }            num[i] = temp;        }    }}int main(){    int t;    getPrime();    scanf("%d",&t);    while(t--)    {        res = -1;        scanf("%d %d",&s,&e);        BFS();        if(res == -1)            printf("Impossible\n");        else            printf("%d\n",res);    }    return 0;}