假設我們有一個 h x w 的網格。網格在一個名為 'initGrid' 的二維數組中表示,其中網格中的每個單元格都以 '#' 或 '.' 表示。 '#' 表示網格中有障礙物,'.' 表示該單元格上有一條路徑。現在,一個機器人被放置在網格上的一個單元格 'c' 上,該單元格具有行號 x 和列號 y。機器人必須從一個具有行號 p 和列號 q 的單元格 'd' 移動到另一個單元格。單元格座標 c 和 d 都以整數對的形式給出。現在,機器人可以按以下方式從一個單元格移動到另一個單元格:
如果機器人想要移動到的單元格位於目前單元格的垂直或水平相鄰位置,機器人可以直接從一個單元格走到另一個單元格。
機器人可以跳到以其目前位置為中心的 5x5 區域中的任何單元格。
機器人只能移動到不包含障礙物的網格中的另一個單元格。機器人也不能離開網格。
我們需要找出機器人到達目標所需的跳數。
因此,若輸入為h = 4,w = 4,c = {2, 1},d = {4, 4},initGrid = {"#...", ".##.", " ...#", "..#."},那麼輸出將為1。機器人只需要一次跳躍就可以到達目的地。
為了解決這個問題,我們將按照以下步驟進行:
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])
#讓我們看下面的實作以獲得更好的理解−
#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; }
4, 4, {2, 1}, {4, 4}, {"#...", ".##.", "...#", "..#."}
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以上是C++程式用來找出機器人在網格中到達特定單元所需的跳躍次數的詳細內容。更多資訊請關注PHP中文網其他相關文章!