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  • nyoj 592 spiral grid

    spiral grid

    时间限制:2000 ms  |  内存限制:65535 KB

    难度:4

    描述

    Xiaod 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. In addition, traveling from a prime number is disallowed, either. 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.

     

    输入

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

    输出

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

    样例输入

    1 4

    9 32

    10 12

    样例输出

    Case 1: 1

    Case 2: 7

    Case 3: impossible

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<queue>
    using namespace std;
    int map[105][105]={0};
    int mm=10000,n=100;
    int fir,end,loop[105][105];
    int prime[11000]={1,1,0};
    int sx[]={0,0,1,-1},zy[]={1,-1,0,0};
    void Build_map() //建立地图
    {
        int i,j,k;
        for(i=1;i<=(n+1)/2;i++)
        {
            for(j=1;j<=n-2*i+2;j++)
                map[i][j+i-1]=mm--;
            for(j=1;j<=n-2*i;j++)
               map[j+i][n-i+1]=mm--;
            for(j=n-i+1;j>=i;j--)
               map[n-i+1][j]=mm--;
            for(j=n-i;j>=i+1;j--)
               map[j][i]=mm--;
        }
    }
    void printf_map()//打印地图
    {
        int i,j;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
              printf("%d ",map[i][j]);
            printf("
    ");
        }
    
    }
    void judge_prime()  //筛选法打个素数表
    {
        int i,j;
        for(i=2;i<=10000;i++)
        {
            if(prime[i]==0)
            {
                for(j=i+i;j<=10000;j+=i)
                  prime[j]=1;
            }
        }
    }
    struct coordinate
    {
        int x;
        int y;
        int step;
    }t1;
    void find() //找到 fir 的坐标,存在 t1 中
    {
        int i,j;
        for(i=1;i<=n;i++)
          for(j=1;j<=n;j++)
          {
              if(map[i][j]==fir)
              {
                  t1.x=i;
                  t1.y=j;
                  loop[i][j]=1;
                  t1.step=0;
                  return;
              }
          }
    }
    int bfs()
    {
        int i,j,step,x,y;
        queue<coordinate> Q;
        Q.push(t1);
        while(!Q.empty())
        {
            i=Q.front().x;
            j=Q.front().y;
            step=Q.front().step;
            Q.pop();
            if(map[i][j]==end)
              return step;
            for(int a=0;a<4;a++)
            {
                x=i+sx[a];
                y=j+zy[a];
                if(map[x][y]!=0&&prime[map[x][y]]==1&&loop[x][y]==0)
                {
                    coordinate t2={x,y,step+1};
                    Q.push(t2);
                    loop[x][y]=1;
                }
            }
        }
        return 0;
    }
    int main()
    {
        int i,j,nn=0;
        Build_map();
        judge_prime();
        //printf_map();
        while(~scanf("%d%d",&fir,&end))
        {
            nn++;
            memset(loop,0,sizeof(loop));
            find();
            int ans=bfs();
            if(ans)
                printf("Case %d: %d
    ",nn,ans);
            else
                printf("Case %d: impossible
    ",nn);
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangliu/p/7053055.html
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