A、牛牛爱奇数

class Solution {
public:
    /**
     * 返回一个数，代表让这些数都变成奇数的最少的操作次数
     * @param n int整型 代表一共有多少数
     * @param a int整型vector 代表n个数字的值
     * @return int整型
     */
    int solve(int n, vector<int>& a) {
        // write code here
        map<int, int> mp;
        for (auto it : a)    ++mp[it];
        int cnt = 0;
        for (auto i = mp.rbegin(); i != mp.rend(); ++i) {
            int num = i->first, val = i->second;
            if (num & 1)    continue;
            else {
                int tmp = num >> 1;
                if (!(tmp & 1))    mp[tmp] += val;
                ++cnt;
            }
        }
        return cnt;
    }
};

B、牛牛摆放花

const int N = 1e5 + 7;
int a[N];
int b[N];
class Solution {
public:
    /**
     * ​返回按照这些花排成一个圆的序列中最小的“丑陋度”
     * @param n int整型 花的数量
     * @param array int整型vector 花的高度数组
     * @return int整型
     */
    int solve(int n, vector<int>& array) {
        // write code here
        int cnt = 0;
        for (auto it : array)    a[++cnt] = it;
        sort(a + 1, a + 1 + cnt);
        int l = 1, r = cnt, op = 1, tmp = 1;
        while (tmp <= n) {
            if (op)    b[l++] = a[tmp++];
            else b[r--] = a[tmp++];
            op ^= 1;
        }
        b[0] = b[cnt];
        int mini = -1e9;
        for (int i = 1; i <= cnt; ++i)
            mini = max(mini, abs(b[i] - b[i - 1]));
        return mini;
    }
};

C、星球游戏

typedef long long ll;

const int N = 1e5 + 7; //节点数
const int M = 2e5 + 7; //路径数
const ll INF = 0x3f3f3f3f;
int u[M], v[M], w[M];
ll d1[N]; //d1正向图
struct Node {
    int v;
    ll cost;
    bool operator < (Node b)const {
        return cost > b.cost;
    }
};
vector<Node> e[N];
int n, m, st, ed;
ll ans;

void dijkstra(int s, ll d[]) {
    priority_queue<Node>  pq;
    while (pq.size())    pq.pop();
    fill(d, d + N, INF);
    d[s] = 0;
    pq.push(Node{ s,0 });
    while (!pq.empty()) {
        Node    x = pq.top();
        pq.pop();
        if (d[x.v] < x.cost)    continue; //原先这个节点花费更少 跳过
        for (auto it : e[x.v]) {
            if (d[it.v] > x.cost + it.cost) {
                d[it.v] = x.cost + it.cost;
                pq.push(Node{ it.v,d[it.v] });
            }
        }
    }
}
class Solution {
public:
    /**
     *
     * @param niuniu int整型vector 牛牛占领的p个星球的编号
     * @param niumei int整型vector 牛妹占领的q个星球的编号
     * @param path int整型vector<vector<>> m条隧道，每条隧道有三个数分别是ui,vi,wi。ui,vi分别是隧道的两边星球的编号，wi是它们之间的距离
     * @param nn int整型 星球个数n
     * @return int整型
     */
    int Length(vector<int>& niuniu, vector<int>& niumei, vector<vector<int> >& path, int nn) {
        // write code here
        for (int i = 0; i <= nn; ++i)    e[i].clear();
        for (auto it : niuniu)    e[0].push_back({ it,0 });
        for (auto it : path) {
            int u = it[0], v = it[1], w = it[2];
            e[u].push_back(Node{ v, w });
            e[v].push_back(Node{ u, w });
        }
        st = 0;
        ans = INF;
        dijkstra(st, d1);
        for (auto it : niumei) {
            ans = min(ans, d1[it]);
        }
        if (ans == INF)    return -1;
        return ans;
    }
};