当前位置：首页 > news >正文

【今日三题】ISBN号码(模拟) / kotori和迷宫(BFS最短路) / 矩阵最长递增路径(dfs)

news 2025/5/9 6:27:23

⭐️个人主页：@小羊 ⭐️所属专栏：每日两三题很荣幸您能阅读我的文章，诚请评论指点，欢迎欢迎 ~

动图描述

ISBN号码(模拟)

ISBN号码

在这里插入图片描述

#include <iostream>
#include <string>
using namespace std;string s;
int sum, cnt = 1, n;int main()
{cin >> s;n = s.size();for (int i = 0; i < n - 1; i++){if (s[i] != '-'){sum += (s[i] - '0') * cnt;cnt++;}}sum %= 11;if (sum == (s[n - 1] - '0') || (sum == 10 && s[n - 1] == 'X')){cout << "Right" << endl;}else {s[n - 1] = (sum == 10 ? 'X' : (sum + '0'));cout << s << endl;}return 0;
}

kotori和迷宫(BFS最短路)

kotori和迷宫

在这里插入图片描述

#include <iostream>
#include <queue>
using namespace std;int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
const int N = 31;
bool used[N][N];
char arr[N][N];
using pii = pair<int, int>;
queue<pii> q;
int n, m, cnt, dist, step;int main()
{cin >> n >> m;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){cin >> arr[i][j];if (arr[i][j] == 'k'){q.push({i, j});used[i][j] = true;}}}while (q.size()){step++;int sz = q.size();while (sz--){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < n && y >= 0 && y < m && !used[x][y] && arr[x][y] != '*'){if (arr[x][y] == 'e'){if (cnt == 0) dist = step;cnt++;}else if (arr[x][y] == '.'){q.push({x, y});}used[x][y] = true;}}}}if (cnt > 0) cout << cnt << " " << dist << endl;else cout << -1 << endl;return 0;
}

矩阵最长递增路径(dfs)

矩阵最长递增路径

在这里插入图片描述

class Solution {int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};bool used[1001][1001] = {};int n, m;
public:int solve(vector<vector<int> >& matrix) {n = matrix.size(), m = matrix[0].size();int len = 0;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){len = max(len, dfs(matrix, i, j));}}return len;}int dfs(const vector<vector<int>> &matrix, int i, int j){int len = 1;for (int k = 0; k < 4; k++){int x = i + dx[k], y = j + dy[k];if (x >= 0 && x < n && y >= 0 && y < m && !used[x][y] && matrix[x][y] > matrix[i][j]){used[x][y] = true;len = max(len, dfs(matrix, x, y) + 1);used[x][y] = false;}}return len;}
};