简单BFS POJ 3126 Prime Path

题目

题目大意就是输入一个T然后输入一个素数，再输入另一个数，输入第一个数变成另一个所需要的最少步数。

代码：

View Code 
  1 #include <stdio.h>
  2 #include <iostream>
  3 #include <string.h>
  4 
  5 using namespace std;
  6 
  7 int vis[10005];
  8 int prim[10005],leap;
  9 struct node
 10 {
 11     int n[5],step;
 12     int num;
 13 }q[4*10005];
 14 void make_prim()
 15 {
 16     memset(prim,0,sizeof(prim));
 17     int flag = 1,i,j;
 18     for(i = 1001;i <= 9999;i += 2)
 19     {
 20         flag = 1;
 21         for(j = 2;j <= i/2;j++)
 22         {
 23             if(i%j == 0)
 24             {
 25                 flag = 0;
 26                 break;
 27             }
 28         }
 29         prim[i] = flag;
 30 
 31     }
 32 }
 33 int num,f,r,target;
 34 void get()
 35 {
 36     int n;
 37     n = q[f].num;
 38 
 39     q[f].n[4] = n%10;
 40     n /= 10;
 41     q[f].n[3] = n%10;
 42     n /= 10;
 43     q[f].n[2] = n%10;
 44     n /= 10;
 45     q[f].n[1] = n;
 46 }
 47 int getsum(struct node temp)
 48 {
 49     return temp.n[1]*1000+temp.n[2]*100+temp.n[3]*10+temp.n[4];
 50 }
 51 void bfs()
 52 {
 53     int i,j;
 54     memset(vis,0,sizeof(vis));
 55     f = r = 0;
 56     q[r].num = num;
 57     q[r].step = 0;
 58     get();
 59     r++;
 60     vis[num] = 1;
 61     leap = 0;
 62     while(f < r)
 63     {
 64         struct node temp,now;
 65         get();
 66         now  = q[f];
 67         now.step = q[f].step+1;
 68         f++;
 69         for(i = 0;i < 10;i++)
 70         {
 71             for(j = 1;j < 5;j++)
 72             {
 73                 temp = now;
 74                 if(i != temp.n[j])
 75                 {
 76                     temp.n[j] = i;
 77                     temp.num = getsum(temp);
 78 
 79                     if(!vis[temp.num] && prim[temp.num])
 80                     {
 81                         vis[temp.num] = 1;
 82                         q[r] = temp;
 83                         if(temp.num == target)
 84                         {
 85                             leap = 1;
 86                             break;
 87                         }
 88                         r++;
 89                     }
 90                 }
 91             }
 92             if(leap)
 93             break;
 94         }
 95         if(leap)
 96         break;
 97     }
 98     return ;
 99 }
100 int main()
101 {
102     int t;
103     make_prim();
104     scanf("%d",&t);
105 
106     while(t--)
107     {
108         scanf("%d %d",&num,&target);
109         if(num == target)
110         {
111             puts("0");
112             continue;
113         }
114 
115         bfs();
116         if(leap)
117         printf("%d\n",q[r].step);
118         else
119         printf("Impossible\n");
120 
121     }
122 
123     return 0;
124 }
posted @ 2012-09-07 17:29  某某。  阅读(221)  评论(0)  编辑  收藏  举报