题目链接

我们看到这道题，一下子就可以想到用模拟来做，但是！地图存不下去，这是一个严重的问题。

怎么办呢？

方法一：

我们可以用不同的数字来代替字符，再用short存。

CODE：

#include <iostream>
#include <vector>
#include <string>using namespace std;int main() {// 读取输入int n, m, x;cin >> n >> m >> x;// 读取网格vector<vector<short>> grid(n, vector<short>(m));int start_i = 0, start_j = 0; // 初始位置for (int i = 0; i < n; ++i) {string row;cin >> row;for (int j = 0; j < m; ++j) {if (row[j] == '#') {start_i = i;start_j = j;grid[i][j] = 0; // 起点视为空地} else if (row[j] == 'X') {grid[i][j] = 1; // 1 表示障碍物} else {grid[i][j] = 0; // 0 表示空地}}}// 读取移动方向vector<short> opts(x);for (int i = 0; i < x; ++i) {cin >> opts[i];}// 模拟游走int current_i = start_i, current_j = start_j; // 当前位置for (int i = 0; i < x; ++i) {int opt = opts[i];int next_i = current_i, next_j = current_j;// 计算下一步的坐标if (opt == 1) {next_i--; // 上} else if (opt == 2) {next_i++; // 下} else if (opt == 3) {next_j--; // 左} else if (opt == 4) {next_j++; // 右}// 检查是否越界或撞到障碍物if (next_i < 0 || next_i >= n || next_j < 0 || next_j >= m || grid[next_i][next_j] == 1) {cout << "ZCS is die!" << endl;return 0;}// 更新当前位置current_i = next_i;current_j = next_j;}// 输出最终坐标cout << current_i - start_i << " " << current_j - start_j << endl;return 0;
}

方法二：

我们把障碍用vector存下来。

CODE：

#include <iostream>
using namespace std;// 定义方向偏移量，分别对应上、下、左、右
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};int main() {short n, m, x;// 读取教室的长、宽和步数cin >> n >> m >> x;short startX, startY;char cell;// 读取地图并确定起点坐标for (short i = 0; i < n; ++i) {for (short j = 0; j < m; ++j) {cin >> cell;if (cell == '#') {startX = j;startY = i;}}}short currentX = startX;short currentY = startY;for (short i = 0; i < x; ++i) {short opt;cin >> opt;// 方向选项减 1 以匹配偏移量数组的索引opt--;short newX = currentX + dx[opt];short newY = currentY + dy[opt];// 动态检查新位置是否越界或者遇到障碍bool isInvalid = false;if (newX < 0 || newX >= m || newY < 0 || newY >= n) {isInvalid = true;} else {// 重新读取地图中对应位置的信息short pos = newY * m + newX;short currentRow = 0;cin.clear();cin.seekg(0, ios::beg);for (short k = 0; k < n; ++k) {for (short l = 0; l < m; ++l) {cin >> cell;if (k * m + l == pos) {if (cell == 'X') {isInvalid = true;}break;}}if (isInvalid) {break;}}}if (isInvalid) {cout << "ZCS is die!" << endl;return 0;}currentX = newX;currentY = newY;}// 如果没有撞到障碍，输出最终坐标cout << "(" << currentX - startX << ", " << currentY - startY << ")" << endl;return 0;
}    

注：有可能过不了。 

方法三：

我们可以把经历过的坐标记录下来，用hash去重，并且看是否与某个障碍物的坐标重合了。

CODE:

#include <iostream>
#include <vector>
#include <string>
#include <unordered_set>using namespace std;// 定义一个哈希函数用于存储坐标
struct CoordinateHash {size_t operator()(const pair<int, int>& p) const {return static_cast<size_t>(p.first) * 100000 + p.second;}
};int main() {// 读取输入int n, m, x;cin >> n >> m >> x;// 读取网格vector<string> grid(n);int start_i = 0, start_j = 0; // 初始位置for (int i = 0; i < n; ++i) {cin >> grid[i];for (int j = 0; j < m; ++j) {if (grid[i][j] == '#') {start_i = i;start_j = j;}}}// 读取移动方向vector<int> opts(x);for (int i = 0; i < x; ++i) {cin >> opts[i];}// 记录障碍物的坐标unordered_set<pair<int, int>, CoordinateHash> obstacles;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (grid[i][j] == 'X') {obstacles.insert({i - start_i, j - start_j});}}}// 模拟游走int current_i = 0, current_j = 0; // 当前位置（相对于起点）unordered_set<pair<int, int>, CoordinateHash> visited; // 记录走过的坐标visited.insert({current_i, current_j});for (int i = 0; i < x; ++i) {int opt = opts[i];int next_i = current_i, next_j = current_j;// 计算下一步的坐标if (opt == 1) {next_i--; // 上} else if (opt == 2) {next_i++; // 下} else if (opt == 3) {next_j--; // 左} else if (opt == 4) {next_j++; // 右}// 检查是否撞到障碍物if (obstacles.find({next_i, next_j}) != obstacles.end()) {cout << "ZCS is die!" << endl;return 0;}// 检查是否越界if (next_i < -start_i || next_i >= n - start_i || next_j < -start_j || next_j >= m - start_j) {cout << "ZCS is die!" << endl;return 0;}// 更新当前位置current_i = next_i;current_j = next_j;visited.insert({current_i, current_j});}// 输出最终坐标cout << current_i << " " << current_j << endl;return 0;
}

多么简单的一道题。