六边形格子地图的基本实现

基本思路

这是一张大小为 21 × 15 的六边形格子地图

我们将左上角的格子定义为 (1, 1)
则部分地图的坐标如下

易知:
若起点为(x, y)
当y为奇数时
可以前往的方向为
(x-1, y-1) (x, y-1) (x+1, y) (x, y+1) (x-1, y+1) (x-1, y)
当y为偶数时
可以前往的方向为
(x-1, y) (x, y-1) (x+1, y-1) (x+1, y) (x+1, y+1) (x, y+1)

然后写个bfs就能判断能到达的地方了

实现代码

#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int max_x = 21, max_y = 15;
struct cell
{bool Is_Block = false; //有无障碍
}hex_map[max_x + 1][max_y + 1];
int to[2][6][2] = { -1,0, 0,-1, 1,-1, 1,0, 1,1, 0,1, -1,-1, 0,-1, 1,0, 0,1, -1,1, -1,0 };
int vis[max_x + 1][max_y + 1];
bool judge(int x, int y)
{if (x == max_x && y % 2 == 0) return false;if (x <= 0 || x > max_x || y <= 0 || y > max_y)return false;if (vis[x][y]) return false;return !hex_map[x][y].Is_Block;
}
void bfs(int X, int Y, int Range)
{int now_x, now_y, i, new_x, new_y;queue<pair<int, int>> que;vis[X][Y] = 1;que.push(make_pair(X, Y));while (!que.empty()){now_x = que.front().first;now_y = que.front().second;que.pop();if (vis[now_x][now_y] == Range + 1) continue;for (i = 0; i < 6; i++){new_x = now_x + to[now_y % 2][i][0];new_y = now_y + to[now_y % 2][i][1];if (judge(new_x, new_y)){vis[new_x][new_y] = vis[now_x][now_y] + 1;que.push(make_pair(new_x, new_y));}}}
}
int main(void)
{int temp, x, y, r;for (int i = 1; i <= max_y; i++){for (int i2 = 1; i2 <= max_x + i % 2 - 1; i2++){scanf("%d", &temp);if (temp)hex_map[i2][i].Is_Block = true;}}scanf("%d %d %d", &x, &y, &r);bfs(x, y, r);for (int i = 1; i <= max_y; i++){for (int i2 = 1; i2 <= max_x + i % 2 - 1; i2++){printf("%2d", vis[i2][i]);}putchar('\n');if ((i % 2)) putchar(' ');}
}

运行样例

输入：
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 3 6 // 起点的x y 可以行动的范围
运行结果：
1为起点 0为不可到达

输入2：
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 4 8