C++程序:计算机器人到达网格中特定单元格所需的跳跃次数


假设我们有一个h x w维度的网格。该网格用一个名为‘initGrid’的二维数组表示,其中网格中的每个单元格都用‘#’或‘.’表示。‘#’表示网格中存在障碍物,‘.’表示该单元格可以通过。现在,一个机器人放置在网格上的单元格'c'上,该单元格的行号为x,列号为y。机器人必须移动到另一个单元格'd',其行号为p,列号为q。单元格坐标c和d都以整数对的形式提供给我们。现在,机器人可以通过以下方式从一个单元格移动到另一个单元格:

  • 如果机器人想要移动到的单元格位于其当前所在单元格的垂直或水平相邻位置,则机器人可以从一个单元格走到另一个单元格。

  • 机器人可以跳到以其当前所在单元格为中心的5×5区域内的任何单元格。

  • 只有当目标单元格不包含障碍物时,机器人才能移动到网格中的另一个单元格。机器人也不能离开网格。

我们必须找出机器人到达目的地所需的跳跃次数。

因此,如果输入类似于h = 4,w = 4,c = {2, 1},d = {4, 4},initGrid = {"#...", ".##.", "...#", "..#."},则输出为1。机器人只需要一次跳跃即可到达目的地。

为了解决这个问题,我们将遵循以下步骤:

N:= 100
Define intger pairs s and t.
Define an array grid of size: N.
Define an array dst of size: N x N.
Define a struct node that contains integer values a, b, and e.
Define a function check(), this will take a, b,
   return a >= 0 AND a < h AND b >= 0 AND b < w
Define a function bfs(), this will take a, b,
   for initialize i := 0, when i < h, update (increase i by 1), do:
   for initialize j := 0, when j < w, update (increase j by 1), do:
      dst[i, j] := infinity
   dst[a, b] := 0
   Define one deque doubleq
   Insert a node containing values {a, b, and dst[a, b]} at the end of doubleq
   while (not doubleq is empty), do:
      nd := first element of doubleq
      if e value of nd > dst[a value of nd, b value of nd], then:
         Ignore the following part, skip to the next iteration
   for initialize diffx := -2, when diffx <= 2, update (increase diffx by 1), do:
   for initialize diffy := -2, when diffy <= 2, update (increase diffy by 1), do:
      tm := |diffx + |diffy||
      nx := a value of nd + diffx, ny = b value of nd + diffy
      if check(nx, ny) and grid[nx, ny] is same as '.', then:
         w := (if tm > 1, then 1, otherwise 0)
         if dst[a value of nd, b value of nd] + w < dst[nx, ny], then:
            dst[nx, ny] := dst[a value of nd, b value of nd] + w
            if w is same as 0, then:
               insert node containing values ({nx, ny, dst[nx, ny]}) at the beginning of doubleq.
          Otherwise
insert node containing values ({nx, ny, dst[nx, ny]}) at the end of doubleq.
s := c
t := d
(decrease first value of s by 1)
(decrease second value of s by 1)
(decrease first value of t by 1)
(decrease second value of t by 1)
for initialize i := 0, when i < h, update (increase i by 1), do:
grid[i] := initGrid[i]
bfs(first value of s, second value of s)
print(if dst[first value of t, second value of t] is same as infinity, then -1, otherwise dst[first value of t, second value of t])

示例

让我们看看下面的实现以更好地理解:

Open Compiler
#include <bits/stdc++.h> using namespace std; const int INF = 1e9; #define N 100 int h, w; pair<int, int> s, t; string grid[N]; int dst[N][N]; struct node { int a, b, e; }; bool check(int a, int b) { return a >= 0 && a < h && b >= 0 && b < w; } void bfs(int a, int b) { for (int i = 0; i < h; i++) { for (int j = 0; j < w; j++) dst[i][j] = INF; } dst[a][b] = 0; deque<node> doubleq; doubleq.push_back({a, b, dst[a][b]}); while (!doubleq.empty()) { node nd = doubleq.front(); doubleq.pop_front(); if (nd.e > dst[nd.a][nd.b]) continue; for (int diffx = -2; diffx <= 2; diffx++) { for (int diffy = -2; diffy <= 2; diffy++) { int tm = abs(diffx) + abs(diffy); int nx = nd.a + diffx, ny = nd.b + diffy; if (check(nx, ny) && grid[nx][ny] == '.') { int w = (tm > 1) ? 1 : 0; if (dst[nd.a][nd.b] + w < dst[nx][ny]) { dst[nx][ny] = dst[nd.a][nd.b] + w; if (w == 0) doubleq.push_front({nx, ny, dst[nx][ny]}); else doubleq.push_back({nx, ny, dst[nx][ny]}); } } } } } } void solve(pair<int,int> c, pair<int, int> d, string initGrid[]){ s = c; t = d; s.first--, s.second--, t.first--, t.second--; for(int i = 0; i < h; i++) grid[i] = initGrid[i]; bfs(s.first, s.second); cout << (dst[t.first][t.second] == INF ? -1 : dst[t.first][t.second]) << '\n'; } int main() { h = 4, w = 4; pair<int,int> c = {2, 1}, d = {4, 4}; string initGrid[] = {"#...", ".##.", "...#", "..#."}; solve(c, d, initGrid); return 0; }

输入

4, 4, {2, 1}, {4, 4}, {"#...", ".##.", "...#", "..#."}

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输出

1

更新于:2022年2月25日

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