【今日三题】旋转字符串 / 合并k个已排序的链表 / 滑雪(dfs)

class Solution {
public:
    bool solve(string A, string B) {
    	if (A.size() != B.size()) return false;
        for (int i = 0; i < A.size(); i++)
        {
            if (A.substr(i) + A.substr(0, i) == B) return true;
        }
        return false;
    }
};

class Solution {
public:
   bool solve(string A, string B) {
       if (A.size() != B.size()) return false;
       return (A + A).find(B) != string::npos;
   }
};

class Solution {
public:
    struct cmp{
        bool operator()(ListNode* l1, ListNode* l2)
        {
            return l1->val > l2->val;
        }
    };
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, cmp> pq;
        for (auto& lt : lists) 
        {
            if (lt) pq.push(lt);
        }
        ListNode newnode(0);
        ListNode* tail(&newnode);
        while (pq.size())
        {
            tail->next = pq.top();
            pq.pop();
            tail = tail->next;
            if (tail->next) pq.push(tail->next);
        }
        return newnode.next;
    }
};

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        return merge(lists, 0, lists.size() - 1);
    }
    ListNode* merge(vector<ListNode*>& lt, int l, int r)
    {
        if (l > r) return nullptr;
        if (l == r) return lt[l]; // 关键
        int mid = (l + r) >> 1;
        return dfs(merge(lt, l, mid), merge(lt, mid + 1, r));
    }
    ListNode* dfs(ListNode* l1, ListNode* l2)
    {
        if (l1 == nullptr) return l2;
        if (l2 == nullptr) return l1;
        if (l1->val < l2->val)
        {
            l1->next = dfs(l1->next, l2);
            return l1;
        }
        else 
        {
            l2->next = dfs(l1, l2->next);
            return l2;
        }
    }
};

#include <iostream>
using namespace std;

int n, m;
int arr[110][110];
bool used[110][110];
int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, -1, 1};
int res;

int dfs(int i, int j)
{
    used[i][j] = true;
    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] && arr[x][y] < arr[i][j])
        {
            len = max(len, dfs(x, y) + 1);
        }
    }
    used[i][j] = false;
    return len;
}

int main() 
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            cin >> arr[i][j];
        } 
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            res = max(res, dfs(i, j));
        }
    }
    cout << res << endl;
    return 0;
}