洛谷 P3605 [USACO17JAN]Promotion Counting晋升者计数

题目描述

The cows have once again tried to form a startup company, failing to remember from past experience that cows make terrible managers!

The cows, conveniently numbered $1 \ldots N\left( 1\leq N\leq 100,000\right)$ organize the company as a tree, with cow 1 as the president (the root of the tree). Each cow except the president has a single manager (its "parent" in the tree). Each cow $i$ has a distinct proficiency rating, $p(i)$, which describes how good she is at her job. If cow $i$ is an ancestor (e.g., a manager of a manager of a manager) of cow $j$, then we say $j$ is a subordinate of $i$.

Unfortunately, the cows find that it is often the case that a manager has less proficiency than several of her subordinates, in which case the manager should consider promoting some of her subordinates. Your task is to help the cows figure out when this is happening. For each cow $i$ in the company, please count the number of subordinates $j$ where $p(j)>p(i)$.

奶牛们又一次试图创建一家创业公司，还是没有从过去的经验中吸取教训--牛是可怕的管理者！

为了方便，把奶牛从 $1 \ldots N\left( 1\leq N\leq 100,000\right)$ 编号，把公司组织成一棵树，1 号奶牛作为总裁（这棵树的根节点）。除了总裁以外的每头奶牛都有一个单独的上司（它在树上的 “双亲结点”）。所有的第 $i$ 头牛都有一个不同的能力指数 $p(i)$，描述了她对其工作的擅长程度。如果奶牛 $i$ 是奶牛 $j$ 的祖先节点（例如，上司的上司的上司），那么我们我们把奶牛 $j$ 叫做 $i$ 的下属。

不幸地是，奶牛们发现经常发生一个上司比她的一些下属能力低的情况，在这种情况下，上司应当考虑晋升她的一些下属。你的任务是帮助奶牛弄清楚这是什么时候发生的。简而言之，对于公司的中的每一头奶牛 $i$，请计算其下属 $j$ 的数量满足 $p(j)>p(i)$。
输入输出格式
输入格式：

The first line of input contains $N$.

The next NNN lines of input contain the proficiency ratings $p(1) \ldots p(N)$ for the cows. Each is a distinct integer in the range $1 \ldots 1,000,000,000$.

The next $N-1$ lines describe the manager (parent) for cows $2 \ldots N$. Recall that cow 1 has no manager, being the president.

输入的第一行包括一个整数 $N$。

接下来的 $N$ 行包括奶牛们的能力指数 $p(1) \ldots p(N)$. 保证所有数互不相同，在区间 $1 \ldots 10^9$之间。

接下来的 $N−1$ 行描述了奶牛 $2 \ldots N$ 的上司（双亲节点）的编号。再次提醒，1 号奶牛作为总裁，没有上司。

输出格式：

Please print $N$ lines of output. The $i$th line of output should tell the number of subordinates of cow $i$ with higher proficiency than cow $i$.

输出包括 $N$ 行。输出的第 $i$ 行应当给出有多少奶牛 $i$ 的下属比奶牛 $i$ 能力高。

输入输出样例
输入样例#1： 复制

5
804289384
846930887
681692778
714636916
957747794
1
1
2
3

输出样例#1： 复制

2
0
1
0
0

说明

感谢@rushcheyo 的翻译

这题有很多做法。但我目前只会一种。慢慢update。

解法一：dfs序+树状数组。

先求出dfs序和每个节点子树中节点个数，然后把所有节点按权值从大到小排。从大到小插入树状数组，插入之前查询一下dfs序区间和就OK了。

#include <bits/stdc++.h>
#define eb emplace_back
using namespace std;

inline int read() {
    int x = , f = ; char ch = getchar();
    while (ch < '' || ch > '') { if (ch == '-') f = -; ch = getchar(); }
    while (ch >= '' && ch <= '') { x = x *  + ch - ; ch = getchar(); }
    return x * f;
}

const int N = 1e5 + ;
vector<int> G[N];
struct P {
    int u, val;
    bool operator < (const P &rhs) const {
        return val > rhs.val;
    }
} p[N];
int dfn[N], size[N], n, cnt, tree[N];

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

void add(int x) {
    for (int i = x; i <= n; tree[i]++, i += lowbit(i));
}

int query(int x) {
    int res = ;
    for (int i = x; i; res += tree[i], i -= lowbit(i));
    return res;
}

void dfs(int u, int fa) {
    size[u] = ;
    dfn[u] = ++cnt;
    for (auto v : G[u]) {
        if (v == fa) continue;
        dfs(v, u);
        size[u] += size[v];
    }
}

int ans[N];

int main() {
    n = read();
    for (int i = ; i <= n; i++) {
        p[i].val = read();
        p[i].u = i;
    }
    for (int i = ; i <= n; i++) {
        int u = read();
        G[u].eb(i);
    }
    dfs(, );
    sort(p + , p + n + );
    for (int i = ; i <= n; i++) {
        int u = p[i].u;
        ans[u] = query(dfn[u] + size[u] - ) - query(dfn[u]);
        add(dfn[u]);
    }
    for (int i = ; i <= n; i++) printf("%d\n", ans[i]);
    return ;
}

解法二：线段树合并

先离散化，对每个点建一个权值线段树，然后从下往上合并，答案即为当前权值线段树[a[u] + 1, n]的区间和（这玩意好像还挺有趣的？

#include <bits/stdc++.h>
#define eb emplace_back
using namespace std;

inline int read() {
    int x = , f = ; char ch = getchar();
    while (ch < '' || ch > '') { if (ch == '-') f = -; ch = getchar(); }
    while (ch >= '' && ch <= '') { x = x *  + ch - ; ch = getchar(); }
    return x * f;
}

const int N = 1e5 + ;
int rt[N], cnt, n, sum[ * N], rs[N * ], ls[N * ], a[N], ans[N];
vector<int> G[N];
struct P {
    int v, id;
    bool operator < (const P &rhs) const {
        return v < rhs.v;
    }
} p[N];

inline void pushup(int u) {
    sum[u] = sum[ls[u]] + sum[rs[u]];
}

void build(int &u, int l, int r, int x) {
    if (!u) u = ++cnt;
    if (l == r) {
        sum[u]++;
        return;
    }
    int mid = l + r >> ;
    if (x <= mid) build(ls[u], l, mid, x);
    else build(rs[u], mid + , r, x);
    pushup(u);
}

int merge(int u, int v) {
    if (!u || !v) return u + v;
    ls[u] = merge(ls[u], ls[v]);
    rs[u] = merge(rs[u], rs[v]);
    pushup(u);
    return u;
}

int query(int u, int l, int r, int x, int y) {
    if (!u) return ;
    if (x <= l && y >= r) return sum[u];
    int mid = l + r >> ;
    int ans = ;
    if (x <= mid) ans += query(ls[u], l, mid, x, y);
    if (y > mid) ans += query(rs[u], mid + , r, x, y);
    return ans;
}

void dfs(int u) {
    for (auto v : G[u]) {
        dfs(v);
        rt[u] = merge(rt[u], rt[v]);
    }
    ans[u] = query(rt[u], , n, a[u] + , n);
}

int main() {
    n = read();
    for (int i = ; i <= n; i++) {
        p[i].v = read();
        p[i].id = i;
    }
    sort(p + , p + n + );
    for (int i = ; i <= n; i++) {
        a[p[i].id] = i;
        build(rt[p[i].id], , n, i);
    }
    for (int i = ; i <= n; i++) {
        int u = read();
        G[u].eb(i);
    }
    dfs();
    for (int i = ; i <= n; i++) printf("%d\n", ans[i]);
}