目录

1.ISBN号码

2.kotori和迷宫

3.矩阵最长递增路径

---

1.ISBN号码

链接https://www.nowcoder.com/practice/95712f695f27434b9703394c98b78ee5?tpId=290&tqId=39864&ru=/exam/oj

提取题意，模拟一下即可。

#include <iostream>
using namespace std;
int main() {
	string s;
	cin >> s;

	int n = s.size();
	int sum = 0;
	int k = 1;
	for (int i = 0; i < n - 1; ++i)
	{
		if (s[i] != '-')
		{
			sum += (s[i] - '0') * k;
			++k;
		}
	}

	sum %= 11;
	if (sum == 10 && s[n - 1] == 'X' || sum == (s[n - 1] - '0'))
		cout << "Right" << endl;
	else
	{
		for (int i = 0; i < n - 1; ++i)
			cout << s[i];
		if (sum == 10)
			cout << 'X';
		else
			cout << sum;
	}
	return 0;
}

2.kotori和迷宫

链接https://ac.nowcoder.com/acm/problem/50041

BFS / DFS（宽度 / 深度 优先遍历即可）

DFS：（我写的DFS目前还是没找到为什么有测试用例过不去）

// DFS(有测试用例过不了)
#include <iostream>
using namespace std;
const int N = 35;

int n, m;
int cnt = 0;
int sum = 0x3f3f3f3f;
char arr[N][N];
bool vis[N][N] = { false };
int dx[4] = { 0, 0, 1, -1 };
int dy[4] = { 1, -1, 0, 0 };

void DFS(int d, int x, int y)
{
	if (arr[x][y] == 'e')
	{
		cnt++;
		sum = min(sum, d);
		return;
	}

	vis[x][y] = true;
	for (int i = 0; i < 4; ++i)
	{
		int a = x + dx[i];
		int b = y + dy[i];
		if (a >= 1 && a <= n && b >= 1 && b <= m && !vis[a][b] && arr[a][b] != '*')
		{
			DFS(d + 1, a, b);
		}
	}
	vis[x][y] = false;
}

int main()
{
	int x, y;
	cin >> n >> m;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			cin >> arr[i][j];
			if (arr[i][j] == 'k')
			{
				x = i;
				y = j;
			}
		}
	}

	DFS(0, x, y);
	if (cnt == 0)
		cout << -1;
	else
		cout << cnt << ' ' << sum << endl;
	return 0;
}

 BFS：（因为可以保证第一次找到的一定是最近的）

// BFS
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 35;
int x1, y1; // 标记起点位置
int n, m;
char arr[N][N];
int dist[N][N];
queue<pair<int, int>> q;
int dx[4] = { 0, 0, 1, -1 };
int dy[4] = { 1, -1, 0, 0 };
void bfs()
{
	memset(dist, -1, sizeof dist);
	dist[x1][y1] = 0;
	q.push({ x1, y1 });

	while (q.size())
	{
		auto [x2, y2] = q.front();
		q.pop();
		for (int i = 0; i < 4; i++)
		{
			int a = x2 + dx[i], b = y2 + dy[i];
			if (a >= 1 && a <= n && b >= 1 && b <= m && dist[a][b] == -1 &&
				arr[a][b] != '*')
			{
				dist[a][b] = dist[x2][y2] + 1;
				if (arr[a][b] != 'e')
				{
					q.push({ a, b });
				}
			}
		}
	}
}
int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			cin >> arr[i][j];
			if (arr[i][j] == 'k')
			{
				x1 = i, y1 = j;
			}
		}
	}

	bfs();

	int count = 0, ret = 1e9;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			if (arr[i][j] == 'e' && dist[i][j] != -1)
			{
				count++;
				ret = min(ret, dist[i][j]);
			}
		}
	}
	if (count == 0) cout << -1 << endl;
	else cout << count << " " << ret << endl;

	return 0;
}

3.矩阵最长递增路径

链接https://www.nowcoder.com/practice/7a71a88cdf294ce6bdf54c899be967a2?tpId=196&tqId=37184&ru=/exam/oj

深度优先遍历即可（DFS）：

#include <cstdlib>
#include <vector>
class Solution {
public:
	const static int N = 1010;
	bool vis[N][N] = { false };
	int dx[4] = { 0, 0, 1, -1 };
	int dy[4] = { 1, -1, 0, 0 };
	int n, m, ret = 0;
	vector<vector<int>> arr;

	bool Check(int x, int y)
	{
		for (int i = 0; i < 4; ++i)
		{
			int a = x + dx[i];
			int b = y + dy[i];
			if (a >= 0 && a < n && b >= 0 && b < m && arr[x][y] < arr[a][b])
				return true;
		}
		return false;
	}

	void DFS(int x, int y, int d)
	{
		if (!Check(x, y))
		{
			ret = max(ret, d);
			return;
		}

		vis[x][y] = true;
		for (int i = 0; i < 4; ++i)
		{
			int a = x + dx[i];
			int b = y + dy[i];
			if (a >= 0 && a < n && b >= 0 && b < m && !vis[a][b] && arr[x][y] < arr[a][b])
			{
				DFS(a, b, d + 1);
			}
		}
		vis[x][y] = false;
	}

	int solve(vector<vector<int>>& matrix) {
		arr = matrix;
		n = matrix.size();
		m = matrix[0].size();
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < m; ++j)
				DFS(i, j, 0);
		return ret + 1;
	}
};