HDU - 4255 A Famous Grid 【BFS】

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我是靠谱客的博主孤独红酒，最近开发中收集的这篇文章主要介绍HDU - 4255 A Famous Grid 【BFS】，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;

const int MAX = 210; // 方格边长
int MAX2 = MAX*MAX;
int a[MAX+2][MAX+2];
bool isPrime[MAX*MAX+5];
bool vd[MAX+2][MAX+2];

//构造方格
void createGrid(){
    int num = MAX * MAX;
    int top = 1, bottom = MAX, left = 1, right = MAX;
    int x = 1, y = 1; // 左上角(1, 1)是起始点
    while (num >= 1){
        for ( ; y <= right; y++) // 向右
            a[x][y] = num--;
        top++;
        for (x++,y--; x <= bottom; x++) // 向下
            a[x][y] = num--;
        right--;
        for (x--,y--; y >= left; y--) // 向左
            a[x][y] = num--;
        bottom--;
        for (x--,y++; x >= top; x--) // 向上
            a[x][y] = num--;
        x++,y++;
        left++;
    }
}

// 构造素数表
void getPrime(){
    memset(isPrime, 1, sizeof(isPrime));
    isPrime[1] = false;
    for (int i = 2; i <= MAX2; i++){
        if (isPrime[i]){
            for (int j = i+i; j <= MAX2; j+=i)
                isPrime[j] = false;
        }
    }
}

//获取元素坐标
void getLocate(int v, int &x, int &y){
    for (int i = 1; i <= MAX; i++)
        for (int j = 1; j <= MAX; j++)
            if (a[i][j] == v){
                x = i;
                y = j;
                return ;
            }
}

typedef struct{
    int x, y, step;
}node;

// 方向：上、左、下、右
int dir[4][2] = {{-1,0}, {0,-1}, {1,0}, {0,1}};
int bfs(int v, int u){
    if (v == u)
        return 0;
    queue<node> q;
    int vx, vy;
    node tmp, now;
    memset(vd, 0, sizeof(vd));
    getLocate(v, vx, vy);
    tmp.x = vx, tmp.y = vy, tmp.step = 0;
    vd[vx][vy] = true;
    q.push(tmp);
    while (!q.empty()){
        tmp = q.front();
        q.pop();
        for (int i = 0; i < 4; i++){
            now.x = tmp.x + dir[i][0];
            now.y = tmp.y + dir[i][1];
            now.step = tmp.step + 1;
            if (now.x<1 || now.x>MAX || now.y<1 || now.y>MAX || isPrime[a[now.x][now.y]])
                continue;
            if (a[now.x][now.y] == u)
                return now.step;
            if (!vd[now.x][now.y]){
                vd[now.x][now.y] = true;
                q.push(now);
            }
        }
    }
    return -1;
} 

int main()
{
    createGrid();    
    getPrime();
    int u, v;
    int t = 1;
    int ans;
    while (~scanf("%d %d", &v, &u)){
        ans = bfs(v, u);
        if (ans == -1)
            printf("Case %d: impossiblen", t++);
        else
            printf("Case %d: %dn", t++, ans);
    }
    
    return 0;
}