POJ

题目大意：给两个四位的素数 A，B，每次只允许变换一个数字，换完后的四位数也要求是素数，问几步能使 A 换成 B，无法换到输出 Impossible
解题思路：4 位数字，每位数字可以换成 0～9，粗糙看一下是40个入口的BFS。
剪枝：
1、千位不能为 0
2、不能换成自己
3、替换后的数字也要是素数
判定素数这边先打一个素数表

//素数打表void isprime() {    memset(prime, true, sizeof(prime));    for (int i = 2; i <= sqrt(MAXN); i++)         for (int j = i+i; j <= MAXN; j+=i)            prime[j] = false;}

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#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<string.h>#include<queue>#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))const int INF = 0x3f3f3f3f;const int NINF = -INF -1;const int MAXN = 10000+10;using namespace std;bool prime[MAXN], vis[MAXN];int cnt[MAXN];queue<int> q;int T, a, b;bool flag;void isprime() {    memset(prime, true, sizeof(prime));    for (int i = 2; i <= sqrt(MAXN); i++)         for (int j = i+i; j <= MAXN; j+=i)            prime[j] = false;}void bfs(int cur, int tag) {    if (cur == tag) {        flag = true;        return;    }    q.push(cur);    vis[cur] = true;    int dig[4];    while (!q.empty()) {        int now = q.front();        q.pop();//      printf("%d\n", now);        dig[0] = now/1000;        dig[1] = now%1000/100;        dig[2] = now%100/10;        dig[3] = now%10;        for (int i = 0; i < 4; i++) {            int tmp = dig[i];            for (int j = 0; j < 10; j++) {                if (i == 0 && j == 0) continue;                if (j != tmp) {                    dig[i] = j;                    int num = dig[0]*1000+dig[1]*100+dig[2]*10+dig[3];                    if (!vis[num] && prime[num]) {                        cnt[num] = cnt[now] + 1;                        q.push(num);                        vis[num] = true;                    }                    if (num == b) {                        flag = true;                            return;                    }                }            }            dig[i] = tmp;        }    }}int main() {    isprime();    scanf("%d", &T);    while (T--) {        scanf("%d%d", &a, &b);        memset(vis, 0, sizeof(vis));        memset(cnt, 0, sizeof(cnt));        flag = false;        while (!q.empty()) q.pop();        bfs(a, b);        if (flag) printf("%d\n", cnt[b]);        else printf("Impossible\n");    }    return 0;}