第 370 场 LeetCode 周赛题解

A 找到冠军 I

枚举求强于其他所有队的队

class Solution {
public:
    int findChampion(vector<vector<int>> &grid) {
        int n = grid.size();
        int res = 0;
        for (int i = 0; i < n; i++) {
            int t = 0;
            for (int j = 0; j < n; j++)
                if (j != i)
                    t += grid[i][j];
            if (t == n - 1) {
                res = i;
                break;
            }
        }
        return res;
    }
};

---

B 找到冠军 II

计数：若图中入度为 $0$ 的点只有一个则该点为冠军，否则返回 $-1$

class Solution {
public:
    int findChampion(int n, vector<vector<int>> &edges) {
        vector<int> indeg(n);
        for (auto &ei: edges)
            indeg[ei[1]]++;
        vector<int> li;
        for (int i = 0; i < n; i++)
            if (indeg[i] == 0)
                li.push_back(i);
        if (li.size() != 1)
            return -1;
        return li[0];
    }
};

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C 在树上执行操作以后得到的最大分数

动态规划：设 $p[cur][0]$ 为在以 $cur$ 为根的子树上执行若干操作使得该子树是健康的 能得到的最大分数，设 $p[cur][1]$ 为以 $cur$ 为根的子树各节点 $values$ 之和，有状态转移方程：$p[cur][0]=\{\begin{array}{cc}0& ,cur是叶子节点\\ max\{values[cur]+{\sum }_{j是cur的子节点}p[j][0],{\sum }_{j是cur的子节点}p[j][1]\}& ,cur不是叶子节点\end{array}$

class Solution {
public:
    using ll = long long;

    long long maximumScoreAfterOperations(vector<vector<int>> &edges, vector<int> &values) {
        int n = edges.size() + 1;
        vector<int> e[n];
        for (auto &ei: edges) {
            e[ei[0]].push_back(ei[1]);
            e[ei[1]].push_back(ei[0]);
        }
        ll p[n][2];
        memset(p, -1, sizeof(p));//初始化状态
        function<ll(int, int, int)> get = [&](int cur, int type, int par) {//记忆化搜索
            if (p[cur][type] != -1)
                return p[cur][type];
            if (type == 0) {
                if (cur != 0 && e[cur].size() == 1)
                    return p[cur][type] = 0;
                ll r1 = 0, r2 = values[cur];
                for (auto j: e[cur])
                    if (j != par) {
                        r1 += get(j, 1, cur);
                        r2 += get(j, 0, cur);
                    }
                return p[cur][type] = max(r1, r2);
            } else {
                ll r2 = values[cur];
                for (auto j: e[cur])
                    if (j != par)
                        r2 += get(j, 1, cur);
                return p[cur][type] = r2;
            }
        };
        return get(0, 0, 0);
    }
};

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D 平衡子序列的最大和

动态规划 + 离散化 + 树状数组：设数组 $c[i]=nums[i]-i$ ，则 $c$ 数组中非降序子序列的下标序列即为平衡子序列，所有可以对 $c$ 数组进行离散化，然后定义状态 $p[i]$ 为：$c$ 数组中末尾元素为 $i$ 的非降序子序列对应的平衡子序列在 $nums$ 中的最大和，有状态转移方程：$p[i]=max\{\begin{array}{c}nums[i]\\ max\{p[j]|j\le i\}+nums[i]\end{array}$ ，通过树状数组实现其中的前缀极值查询和单点更新

class Solution {
public:
    using ll = long long;
    int N;
    vector<ll> a;

    inline int lowbit(int x) {
        return x & -x;
    }

    void update(int loc, ll val) {// li[loc]=max(li[loc], val);
        for (; loc < N; loc += lowbit(loc))
            a[loc] = max(a[loc], val);
    }

    ll query(int loc) {// max{li[k] | 1<=k<=loc}
        ll res = INT64_MIN;
        for (; loc > 0; loc -= lowbit(loc))
            res = max(res, a[loc]);
        return res;
    }

    long long maxBalancedSubsequenceSum(vector<int> &nums) {
        int n = nums.size();
        vector<int> c(n);
        for (int i = 0; i < n; i++)
            c[i] = nums[i] - i;
        vector<int> t = c;
        sort(t.begin(), t.end());
        t.erase(unique(t.begin(), t.end()), t.end());
        for (int i = 0; i < n; i++)//离散化c
            c[i] = lower_bound(t.begin(), t.end(), c[i]) - t.begin();
        N = n + 1;
        a = vector<ll>(N, INT64_MIN);
        for (int i = 0; i < n; i++) {
            ll t1 = query(c[i] + 1);
            update(c[i] + 1, max(t1, 0LL) + nums[i]);
        }
        return query(n);
    }
};