hdu4255 A Famous Grid（素数表+BFS）

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 172    Accepted Submission(s): 64


Problem Description
Mr. B has recently discovered the grid named "spiral grid".
Construct the grid like the following figure. (The grid is actually infinite. The figure is only a small part of it.)


Considering traveling in it, you are free to any cell containing a composite number or 1, but traveling to any cell containing a prime number is disallowed. You can travel up, down, left or right, but not diagonally. Write a program to find the length of the shortest path between pairs of nonprime numbers, or report it's impossible.
 

 

Input
Each test case is described by a line of input containing two nonprime integer 1 <=x, y<=10,000.
 

 

Output
For each test case, display its case number followed by the length of the shortest path or "impossible" (without quotes) in one line.
 

 

Sample Input
1 4
9 32
10 12
 

 

Sample Output
Case 1: 1
Case 2: 7
Case 3: impossible
 

 

Source
Fudan Local Programming Contest 2012
 
题意：在一个无穷大的蛇形矩阵中，以素数为墙，构成一个图，输入两个合数求这两个合数之间的最短路。
分析：第一步：打素数表；
        第二步：构造蛇形地图；（注意地图大小要大于10000）
        第三步：BFS求解。
 
View Code 
  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<queue>
  5 #define N 40400
  6 #define MAX 210
  7 #define CNT 4000
  8 
  9 using namespace std;
 10 
 11 bool isprime[N];
 12 int prime[CNT],cnt;
 13 int map[MAX][MAX];
 14 bool visited[MAX][MAX];
 15 struct point{int x,y,step;};
 16 
 17 void find_prime()
 18 {
 19     int i,j;
 20     cnt=0;
 21     memset(isprime,true,sizeof(isprime));
 22     isprime[1]=false;
 23     for(i=2;i<N;i++)
 24     {
 25         if(isprime[i])
 26         {
 27             prime[cnt++]=i;
 28             for(j=2*i;j<N;j=j+i) isprime[j]=false;
 29         }
 30     }
 31 }
 32 
 33 void build_map(int p,int q,int s,int c)
 34 {
 35     if(s==0) return ;
 36     int i;
 37     for(i=q;i<q+c;i++)   map[p][i]=s--;
 38     for(i=p+1;i<p+c;i++) map[i][q+c-1]=s--;
 39     for(i=q+c-2;i>q;i--) map[p+c-1][i]=s--;
 40     for(i=p+c-1;i>p;i--) map[i][q]=s--;
 41     build_map(p+1,q+1,s,c-2);
 42 }
 43 
 44 int t_i,t_j;
 45 
 46 void find(int u)
 47 {
 48     int i,j;
 49     for(i=1;i<=200;i++)
 50     {
 51         for(j=1;j<=200;j++)
 52         {
 53             if(u==map[i][j])
 54             {
 55                 t_i=i;t_j=j;
 56                 return ;
 57             }
 58         }
 59     }
 60 }
 61 
 62 int BFS(int u,int v)
 63 {
 64     find(u);
 65     point t,tt;
 66     t.x=t_i;t.y=t_j;t.step=0;
 67     queue<point>Q;
 68     while(!Q.empty()) Q.pop();
 69     Q.push(t);
 70     visited[t.x][t.y]=true;
 71     while(!Q.empty())
 72     {
 73         t=Q.front();Q.pop();
 74         if(map[t.x][t.y]==v) return t.step;
 75         if(t.x-1>=1&&!visited[t.x-1][t.y]&&!isprime[map[t.x-1][t.y]])
 76         {
 77             tt.x=t.x-1;tt.y=t.y;tt.step=t.step+1;
 78             Q.push(tt);
 79             visited[tt.x][tt.y]=true;
 80         }
 81         if(t.x+1<=200&&!visited[t.x+1][t.y]&&!isprime[map[t.x+1][t.y]])
 82         {
 83             tt.x=t.x+1;tt.y=t.y;tt.step=t.step+1;
 84             Q.push(tt);
 85             visited[tt.x][tt.y]=true;
 86         }
 87         if(t.y-1>=1&&!visited[t.x][t.y-1]&&!isprime[map[t.x][t.y-1]])
 88         {
 89             tt.x=t.x;tt.y=t.y-1;tt.step=t.step+1;
 90             Q.push(tt);
 91             visited[tt.x][tt.y]=true;
 92         }
 93         if(t.y+1<=200&&!visited[t.x][t.y+1]&&!isprime[map[t.x][t.y+1]])
 94         {
 95             tt.x=t.x;tt.y=t.y+1;tt.step=t.step+1;
 96             Q.push(tt);
 97             visited[tt.x][tt.y]=true;
 98         }
 99     }
100     return 0;
101 }
102 
103 int main()
104 {
105     int u,v,k=1;
106     memset(map,0,sizeof(map));
107     find_prime();
108     build_map(1,1,40000,200);
109     while(scanf("%d%d",&u,&v)!=EOF)
110     {
111         memset(visited,false,sizeof(visited));
112         int ans;
113         ans=BFS(u,v);
114         if(ans) printf("Case %d: %d\n",k++,ans);
115         else printf("Case %d: impossible\n",k++);
116     }
117     return 0;
118 }

 

posted @ 2012-07-16 23:15  mtry  阅读(668)  评论(1)  编辑  收藏  举报