HDU 4255 A Famous Grid-优快云博客

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1252 Accepted Submission(s): 493

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

以BFS为主的综合题
1.蛇形矩阵
2.筛选法求素数
3.BFS
AC代码

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#define clr(x)  memset(x,0,sizeof(x));
#define MAXN 106
using namespace std;
int a[MAXN+10][MAXN+10],prime[11010],vis[MAXN+10][MAXN+10],pace[MAXN+10][MAXN+10];
int m[4][2]= {{0,1},{1,0},{-1,0},{0,-1}};
void is_prime()
{
    int i,j;
    for(i=2; i<=11000; i++)
    {
        if(prime[i]==0)
            for(j=i+i; j<=11000; j+=i)
                prime[j]=1;
    }
}
int f(int x1,int y1)
{
    if(vis[x1][y1]==0&&x1>=0&&y1<MAXN&&x1<MAXN&&y1>=0&&prime[a[x1][y1]])
        return 1;
    return 0;
}
int bfs(int x,int y)
{
    queue<int> Q,Qx,Qy;
    int i,j,c;
    int ex,ey,fx,fy,tx,ty;
    int flag1=0,flag2=0;
    for(i=0; i<MAXN; i++)
    {
        for(j=0; j<MAXN; j++)
        {
            if(a[i][j]==x) fx=i,fy=j,flag1=1;
            if(a[i][j]==y ) ex=i,ey=j,flag2=1;
        }
        if(flag1&&flag2) break;
    }
    Q.push(x);
    Qx.push(fx);
    Qy.push(fy);
    vis[fx][fy]=1;
    while(!Q.empty())
    {
        c=Q.front();
        Q.pop();
        fx=Qx.front();
        Qx.pop();
        fy=Qy.front();
        Qy.pop();
        if(fx==ex&&fy==ey)
        {
            return  pace[fx][fy] ;
        }
        for(i=0; i<4; i++)
        {
            tx=fx+m[i][0];
            ty=fy+m[i][1];
            if(f(tx,ty))
            {
                vis[tx][ty]=1;
                Q.push(a[tx][ty]);
                Qx.push(tx);
                Qy.push(ty);
                pace[tx][ty]=pace[fx][fy]+1;
            }
        }
    }
    return -1;
}
int main()
{
    int n,x,y,tot =0 ;
    clr(prime);
    clr(a);
    prime[0]=prime[1]=1;
    is_prime();
    n=MAXN;
    clr(a);
    tot = a[x=0][y=0]=n*n;
    while(tot > 1)
    {
        while(y+1<n && !a[x][y+1]) a[x][++y]= --tot;
        while(x+1<n && !a[x+1][y]) a[++x][y] = --tot;
        while(y-1>=0 && !a[x][y-1]) a[x][--y] = --tot;
        while(x-1>=0 && !a[x-1][y]) a[--x][y] = --tot;
    }
    int Case=1;
    while(cin >> x >> y)
    {
        printf("Case %d: ",Case++);
        clr(vis);
        clr(pace);
        int ans=bfs(x,y);
        if(ans==-1) cout << "impossible"<<endl;
        else
            cout << ans << endl;
    }
    return 0;
}