hdu4255 A Famous Grid

最新推荐文章于 2019-03-13 11:33:34 发布

原创最新推荐文章于 2019-03-13 11:33:34 发布 · 1.4k 阅读

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CC 4.0 BY-SA版权

文章标签：

#integer #numbers #input #path #report #struct

本文介绍了一种基于螺旋网格的路径搜索算法实现。该算法通过构建一个包含素数和合数的无限网格，并允许在合数节点间进行移动，目标是最短路径地从一个合数移动到另一个合数。文章提供了完整的C++实现代码，使用了BFS搜索策略来找到最短路径。

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 190 Accepted Submission(s): 71

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

Sample Output

  
   Case 1: 1
Case 2: 7
Case 3: impossible

Source

Fudan Local Programming Contest 2012

简单的搜索题，比赛的时候范围看错了没敢做。多往外拓展几层然后BFS就可以了。

代码很丑……

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;

#define MAXN 200

typedef struct
{
    int x,y;
}Point;

int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
int map[205][205];
int vis[205][205];
queue <Point> q;
Point p[40005];

int Check(int t)
{
    if (t==1) return 1;
    for (int i=2;i*i<=t;i++)
    {
        if (t%i==0) return 1;
    }
    return -1;
}

int BFS(int x,int y)
{
    Point tag,tag1;
    int i;
    if (x==y) return 0;
    vis[p[x].x][p[x].y]=0;
    q.push(p[x]);
    while(!q.empty())
    {
        tag=q.front();
        q.pop();
        for (i=0;i<4;i++)
        {
            tag1.x=tag.x+dx[i];
            tag1.y=tag.y+dy[i];
            if (tag1.x>=MAXN || tag1.x<0 || tag1.y>=MAXN || tag1.y<0 || vis[tag1.x][tag1.y]!=-1 || map[tag1.x][tag1.y]==-1) continue;
            q.push(tag1);
            vis[tag1.x][tag1.y]=vis[tag.x][tag.y]+1;
            if (tag1.x==p[y].x && tag1.y==p[y].y) return vis[tag1.x][tag1.y];
        }
    }
    return -1;
}

int main()
{
    int t,i,j,n,x,y,d,cnt=1,tag;
    d=0;
    t=MAXN*MAXN;
    x=y=0;
    memset(map,0,sizeof(map));
    while(t>0)
    {
        map[x][y]=Check(t);
        p[t].x=x;
        p[t].y=y;
        if (x+dx[d]<0 || x+dx[d]>=MAXN || y+dy[d]<0 || y+dy[d]>=MAXN || map[x+dx[d]][y+dy[d]]!=0)
        {
            d++;
            d%=4;
        }
        x+=dx[d];
        y+=dy[d];
        t--;
    }
    while(scanf("%d%d",&x,&y)!=EOF)
    {
        printf("Case %d: ",cnt++);
        while(!q.empty()) q.pop();
        memset(vis,-1,sizeof(vis));
        tag=BFS(x,y);
        if (tag==-1) printf("impossible\n");
        else printf("%d\n",tag);
    }
    return 0;
}