UPC组队第三场——K: A Famous Grid （BFS+细节）

题目描述
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.

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 Copy
1 4
9 32
10 12
样例输出 Copy
Case 1: 1
Case 2: 7
Case 3: impossible

题意：
给一个螺旋矩阵，每次都只能走非素数的格子，求两个数之间的最短距离。
思路：
首先是考虑如何将这个螺旋矩阵存到数组里，因为在以后需要根据某数的值查找这个数的横纵坐标，在这里用的是map和pair的结合来判断。

int num=106*106,n=106,m=106,j,i;
    for(i=0;i<m/2;i++)
    {
        for(j=i;j<n-i;j++)
          a[i][j]=num,mp[num]={i,j},num--;
        for(j=i+1;j<m-i;j++)
           a[j][n-i-1]=num,mp[num]={j,n-i-1},num--;
            for(j=n-i-2;j>i;j--)
              a[m-i-1][j]=num,mp[num]={m-i-1,j},num--;
            for(j=m-i-1;j>i;j--)
          a[j][i]=num,mp[num]={j,i},num--; 
    }

对于查找两个数之间的距离可以直接用BFS。
本题的坑点就是在存储矩阵的时候如果只打印100 * 100可能就会误判，题目已经说明了是无限大的网格，给出的只是一部分，所以就会存在有一种情况，当打印100*100的矩阵的时候无法到达，当在外圈多打印几圈时，就可以出去再回来，这样是可以到达的。
代码：

#pragma GCC optimize(3)
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll>PLL;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
#define I_int ll
#define x first
#define y second
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char F[200];
inline void out(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
    //cout<<" ";
}
ll ksm(ll a,ll b,ll p){ll res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
const int inf=0x3f3f3f3f,mod=1e9+7;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn=1100,maxm=1e5+7;
const double PI = atan(1.0)*4;
    
int a[maxn][maxn];
map<int,PII>mp;
int res[maxn][maxn];
struct node{
    int x,y,dis;
};
queue<node>q;
bool vis[maxm];//标记非素数
int primer[maxm/10];//存素数
int cnt=0;//记录素数个数
int sx,sy,ex,ey;
int aa[maxn][maxn];
    
void find_primer(){
    vis[1]=1;
    for(int i=2;i<=maxm;i++){
        if(!vis[i])primer[cnt++]=i;//0是素数 
        for(int j=0;j<cnt&&primer[j]*i<=maxm;j++){
            vis[i*primer[j]]=1;
            if(i%primer[j]==0)
            break;
        }
    }
}
    
void init(){
    int num=106*106,n=106,m=106,j,i;
    for(i=0;i<m/2;i++)
    {
        for(j=i;j<n-i;j++)
          a[i][j]=num,mp[num]={i,j},num--;
        for(j=i+1;j<m-i;j++)
           a[j][n-i-1]=num,mp[num]={j,n-i-1},num--;
            for(j=n-i-2;j>i;j--)
              a[m-i-1][j]=num,mp[num]={m-i-1,j},num--;
            for(j=m-i-1;j>i;j--)
          a[j][i]=num,mp[num]={j,i},num--; 
    }
/*  for(i=0;i<m;i++)
     {
       for(j=0;j<n;j++)
          printf("%4d",a[i][j]);
       cout<<endl;
    }*/
}
int nx[4]={0,0,1,-1};
int ny[4]={1,-1,0,0};
    
void bfs(int x,int y){
    while(!q.empty()){
        node t=q.front();q.pop();
        int qx=t.x,qy=t.y;
        ///cout<<qx<<" "<<qy<<endl;
        if(qx==ex&&qy==ey){
            return ;
        }
        for(int i=0;i<4;i++){
            int xx=qx+nx[i],yy=qy+ny[i];
            int tt=a[xx][yy];
            if(!aa[xx][yy]&&vis[tt]&&xx>=0&&xx<106&&yy>=0&&yy<106){
                res[xx][yy]=res[qx][qy]+1;
                q.push({xx,yy,res[xx][yy]});
                aa[xx][yy]=1;
            }
        }
    }
} 
    
int main(){
    find_primer();
    init();///构造矩阵 
  /// cout<<mp[56].first<<" "<<mp[56].second<<endl;
   int st,ed;
    int Case=1;
    while(~scanf("%d%d",&st,&ed)){
        if(vis[st]==0||vis[ed]==0)  
        {
             printf("Case %d: ",Case++);
            puts("impossible");
            continue;
        }
        if(st==ed){
            printf("Case %d: ",Case++);
            puts("0");
            continue;
        }
        sx=mp[st].first,sy=mp[st].second;
        ex=mp[ed].first,ey=mp[ed].second;
    /// cout<<st<<" "<<sx<<" "<<sy<<endl;
    /// cout<<ed<<" "<<ex<<" "<<ey<<endl;
        printf("Case %d: ",Case++);
        memset(res,0,sizeof res);
        memset(aa,0,sizeof aa);
        while(!q.empty()) q.pop();
        q.push({sx,sy,0});
        aa[sx][sy]=1;
        bfs(sx,sy);
        if(res[ex][ey]==0) puts("impossible");
        else out(res[ex][ey]),puts("");
    }
    return 0;
}