BFS POJ 3126 Prime Path

 

题目传送门

  1 /*
  2     题意：从一个数到另外一个数，每次改变一个数字，且每次是素数
  3     BFS：先预处理1000到9999的素数，简单BFS一下。我没输出Impossible都AC，数据有点弱
  4 */
  5 /************************************************
  6 Author        :Running_Time
  7 Created Time  :2015-8-2 15:46:57
  8 File Name     :POJ_3126.cpp
  9 *************************************************/
 10 
 11 #include <cstdio>
 12 #include <algorithm>
 13 #include <iostream>
 14 #include <sstream>
 15 #include <cstring>
 16 #include <cmath>
 17 #include <string>
 18 #include <vector>
 19 #include <queue>
 20 #include <deque>
 21 #include <stack>
 22 #include <list>
 23 #include <map>
 24 #include <set>
 25 #include <bitset>
 26 #include <cstdlib>
 27 #include <ctime>
 28 using namespace std;
 29 
 30 #define lson l, mid, rt << 1
 31 #define rson mid + 1, r, rt << 1 | 1
 32 typedef long long ll;
 33 const int MAXN = 1e4 + 10;
 34 const int INF = 0x3f3f3f3f;
 35 const int MOD = 1e9 + 7;
 36 bool is_prime[MAXN];
 37 bool vis[MAXN];
 38 int x, y, res;
 39 struct Digit    {
 40     int d[5];
 41     int step;
 42 };
 43 
 44 void solve(void)    {
 45     memset (is_prime, true, sizeof (is_prime));
 46     for (int i=1000; i<=9999; ++i)  {
 47         for (int j=2; j<i; ++j)  {
 48             if (i % j == 0) {
 49                 is_prime[i] = false;    break;
 50             }
 51         }
 52     }
 53 }
 54 
 55 int get_num(int *a) {
 56     int ret = 0;
 57     for (int i=1; i<=4; ++i)    {
 58         ret = ret * 10 + a[i];
 59     }
 60     return ret;
 61 }
 62 
 63 void BFS(int u, int v)   {
 64     Digit tmp;
 65     for (int i=4; i>=1; --i)    {
 66         tmp.d[i] = u % 10;  u /= 10;
 67     }
 68     tmp.step = 0;
 69     queue<Digit> Q; Q.push (tmp);   vis[u] = true;
 70     int ans = -1;
 71     while (!Q.empty ()) {
 72         Digit x = Q.front ();   Q.pop ();
 73         int m = get_num (x.d);
 74         if (m == v) {
 75             ans = x.step;   break;
 76         }
 77         for (int i=1; i<=4; ++i)    {
 78             for (int j=0; j<=9; ++j)    {
 79                 if (i == 1 && j == 0)   continue;
 80                 if (x.d[i] != j)    {
 81                     Digit y = x;
 82                     y.d[i] = j; m = get_num (y.d);
 83                     if (is_prime[m] && !vis[m])    {
 84                         vis[m] = true;  y.step++;   Q.push (y);
 85                     }
 86                 }
 87             }
 88         }
 89     }
 90     if (ans == -1)  puts ("Impossible");
 91     else    printf ("%d\n", ans);
 92 }
 93 
 94 int main(void)    {       //POJ 3126 Prime Path
 95     solve ();
 96     int T;  scanf ("%d", &T);
 97     while (T--) {
 98         scanf ("%d%d", &x, &y);
 99         memset (vis, false, sizeof (vis));
100         BFS (x, y);
101     }
102 
103     return 0;
104 }

 

posted @ 2015-08-02 19:56  Running_Time  阅读(146)  评论(0)  编辑  收藏  举报