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BZOJ4756: [Usaco2017 Jan]Promotion Counting(线段树合并)

程序员文章站 2022-04-15 17:20:21
题意 "题目链接" Sol 线段树合并板子题 cpp include using namespace std; const int MAXN = 400000, SS = MAXN 21; inline int read() { char c = getchar(); int x = 0, f = ......

题意

题目链接

sol

线段树合并板子题

#include<bits/stdc++.h>
using namespace std;
const int maxn = 400000, ss = maxn * 21;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, p[maxn], fa[maxn], root[maxn], tot, ans[maxn];
vector<int> v[maxn];
int ls[ss], rs[ss], sum[ss];
void insert(int &k, int l, int r, int p, int v) {
    if(!k) k = ++tot;
    sum[k]++;
    if(l == r) return ;
    int mid = l +  r >> 1;
    if(p <= mid) insert(ls[k], l, mid, p, v);
    else insert(rs[k], mid + 1, r, p, v);
}
int query(int k, int l, int r, int ql, int qr) {
    if(ql <= l && r <= qr) return sum[k];
    int mid = l + r >> 1;
    if(ql > mid) return query(rs[k], mid + 1, r, ql, qr);
    else if(qr <= mid) return query(ls[k], l, mid, ql, qr);
    else return query(ls[k], l, mid, ql, qr) + query(rs[k], mid + 1, r, ql, qr);
}
int merge(int x, int y) {
    if(!x || !y) return x ^ y;
    ls[x] = merge(ls[x], ls[y]);
    rs[x] = merge(rs[x], rs[y]);
    sum[x] += sum[y];
    return x;
}
void dfs(int x) {
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i]; dfs(to);
        root[x] = merge(root[x], root[to]);
    }
    ans[x] = query(root[x], 1, n, p[x] + 1, n);
    insert(root[x], 1, n, p[x], 1);
}
void des() {
    static int date[maxn], num = 0;
    for(int i = 1; i <= n; i++) date[++num] = p[i];
    sort(date + 1, date + num + 1);
    num = unique(date + 1, date + num + 1) - date - 1;
    for(int i = 1; i <= n; i++) p[i] = lower_bound(date + 1, date + n + 1, p[i]) - date;
}
int main() {
    n = read();
    for(int i = 1; i <= n; i++) p[i] = read();
    des();
    for(int i = 2; i <= n; i++) fa[i] = read(), v[fa[i]].push_back(i);
    dfs(1);
    for(int i = 1; i <= n; i++) printf("%d\n", ans[i]);
    return 0;
}
/*
*/