leetcode第 46 场双周赛

Posted 2022-12-12 _luckylight

题目A

https://leetcode-cn.com/problems/longest-nice-substring/
因为 $length\le 100$，我们直接就可以遍历子串然后更新答案。

class Solution 
public:
    bool Check(string s)
    
        unordered_set<int> m;
        for (auto u : s)
            m.insert(u);
        for (auto u : m)
        
            if (u >= 'a' && m.count(u - 32) < 1)    // 97 - 62 = 32
                return false;
            else if (u <= 'Z' && m.count(u + 32) < 1)
                return false;
        
        return true;
    
    string longestNiceSubstring(string s) 
        string ret = "";
        for (int i = 0; i < s.size(); i ++ )
        
            for (int j = i + ret.size(); j < s.size(); j ++ )
            
                if (Check(s.substr(i, j - i + 1)))
                
                    ret = s.substr(i, j - i + 1);
                
            
        
        return ret;
    
;

题目B

https://leetcode-cn.com/problems/form-array-by-concatenating-subarrays-of-another-array/
因为 $length\le 1000$，我们直接就可以直接 $O(N^2)$的复杂度就解决该问题。
直接对$nums$数组遍历，遍历的同时审核是否可以凑出来 $groups$数组

class Solution 
public:
    vector<vector<int> > gps;
    vector<int> nums;
    bool Check(int st, int x)
    
        for (int j = 0; j < gps[x].size(); j ++ )
        
            if (st + j >= nums.size() || gps[x][j] != nums[st + j])
                return false;
        
        return true;
    
    bool canChoose(vector<vector<int>>& groups, vector<int>& _nums) 
        nums = _nums;
        gps = groups;
        int j = 0;
        for (int i = 0; i < nums.size(); i ++ )
        
            if (Check(i, j))
            
                if (j == gps.size() - 1)
                    return true;
                i += gps[j].size() - 1; // 减去 1 是因为后面有 ++
                j ++;
            
        
        return false;
    
;

题目C

https://leetcode-cn.com/problems/map-of-highest-peak/
直接就是一个 bfs 就可以了，主要越界情况，和**vector<vector >**的快速初始化

#define x first
#define y second
typedef pair<int, int> PII;
class Solution 
public: 
    int n, m;
    bool Check(int a, int b)
    
        return a >= 0 && b >= 0 && a < n && b < m;
    

    vector<vector<int>> highestPeak(vector<vector<int>>& isWater) 
        n = isWater.size(), m = isWater[0].size();
        queue<PII> que;
        vector<vector<int> > ret(n, vector<int>(m, -1));
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < m; j ++ )
                if (isWater[i][j] == 1)
                    que.push(PII(i, j)), ret[i][j] = 0;
        
        int dx[4] = 0, 1, 0, -1, dy[4] = 1, 0, -1, 0;
        PII tmp;    int x, y;
        while (que.size())
        
            tmp = que.front();  que.pop();
            for (int i = 0; i < 4; i ++ ) 
                x = tmp.x + dx[i], y = tmp.y + dy[i];
                if (Check(x, y) && ret[x][y] == -1) 
                    ret[x][y] = 1 + ret[tmp.x][tmp.y];
                    que.push(PII(x, y));
                
            
        
        return ret;
    
;

题目D

https://leetcode-cn.com/problems/tree-of-coprimes/
有一个很玄学的问题，就是$TM$
vector copr[55];定义在全局变量那里就会超时，然而类的内部就很快，奇奇怪怪
对于本题的思路就是根据 nums 数字范围太小，我们可以将互质数字进行枚举是否在dfs的路径上。
dfs的路径就是我们根的路径，很方便

const int N = 100010, M = N * 2;
int h[N], e[M], ne[M], idx;
int pos[55];
int depth[N];

class Solution 
public:
    int n;
    vector<int> w;
    vector<int> ret;
    vector<int> copr[55];
    inline int gcd(int x, int y) 
        if (y == 0)
            return x;
        else
            return gcd(y, x % y);
    

    void add(int x, int y) 
        e[idx] = y, ne[idx] = h[x], h[x] = idx ++;
    

    void dfs(int u, int fa) 
        int val = w[u];
        for (auto & x : copr[val])
            if (pos[x] != -1)
                if (ret[u] == -1 || depth[ret[u]] < depth[pos[x]])
                    ret[u] = pos[x];
        int tmp = pos[val];
        pos[val] = u;

        int v;
        for (int i = h[u]; ~i; i = ne[i]) 
            v = e[i];
            if (v == fa)    continue;
            else 
                depth[v] = depth[u] + 1;
                dfs(v, u);
            
        
        pos[val] = tmp;
    
    vector<int> getCoprimes(vector<int>& nums, vector<vector<int>>& edges) 
        for (int i = 1; i <= 50; i ++ )
            for (int j = 1; j <= 50; j ++ )
                if (gcd(i, j) == 1)
                    copr[i].push_back(j);
        
        n = nums.size(), w = nums;
        memset(h, -1, sizeof h), idx = 0;
        memset(pos, -1, sizeof pos);
        memset(depth, 0, sizeof depth);

        for (auto & t : edges)
            add(t[0], t[1]), add(t[1], t[0]);
        
        ret.resize(n, -1);
        depth[0] = 0;
        dfs(0, -1);

        return ret;
    
;