2018.08.27 [Usaco2017 Jan]Promotion Counting（线段树合并）

描述

The cows have once again tried to form a startup company, failing to remember from past experience t hat cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organize t he company as a tree, with cow 1 as the president (the root of the tree). Each cow except the presid ent has a single manager (its “parent” in the tree). Each cow ii has a distinct proficiency rating, p(i), which describes how good she is at her job. If cow ii is an ancestor (e.g., a manager of a man ager of a manager) of cow jj, then we say jj is a subordinate of ii. Unfortunately, the cows find that it is often the case that a manager has less proficiency than seve ral of her subordinates, in which case the manager should consider promoting some of her subordinate s. Your task is to help the cows figure out when this is happening. For each cow ii in the company, please count the number of subordinates jj where p(j)>p(i). n只奶牛构成了一个树形的公司，每个奶牛有一个能力值pi，1号奶牛为树根。问对于每个奶牛来说，它的子树中有几个能力值比它大的。

输入

The first line of input contains N The next N lines of input contain the proficiency ratings p(1)…p(N) for the cows. Each is a distinct integer in the range 1…1,000,000,000 The next N-1 lines describe the manager (parent) for cows 2…N Recall that cow 1 has no manager, being the president. n，表示有几只奶牛 n<=100000 接下来n行为1-n号奶牛的能力值pi 接下来n-1行为2-n号奶牛的经理（树中的父亲）

输出

Please print N lines of output. The ith line of output should tell the number of subordinates of cow ii with higher proficiency than cow i. 共n行，每行输出奶牛i的下属中有几个能力值比i大

样例输入

5

804289384

846930887

681692778

714636916

957747794

1

1

2

3

样例输出

2

0

1

0

0

康复训练ing。。。

这题本蒟蒻用线段树合并水过去了。。。

暂时没想到其它的做法毕竟实力太弱了

代码；

#include<bits/stdc++.h>
#define N 100005
using namespace std;
inline int read(){
    int ans=0;
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
    return ans;
}
inline void write(int x){
    if(x>9)write(x/10);
    putchar((x%10)^48);
}
int ans[N],a[N],b[N],n,sig,first[N],tot=0,cnt=0,rt[N],siz[N*20],son[N*20][2];
struct edge{int v,next;}e[N<<1];
inline void add(int u,int v){e[++cnt].v=v,e[cnt].next=first[u],first[u]=cnt;}
inline void modify(int&p,int l,int r,int k){
    if(!p)p=++tot;
    if(l==r){siz[p]=1;return;}
    int mid=l+r>>1;
    if(k<=mid)modify(son[p][0],l,mid,k);
    else modify(son[p][1],mid+1,r,k);
    siz[p]=siz[son[p][0]]+siz[son[p][1]];
}
inline int merge(int x,int y,int l,int r){
    if(!x||!y)return x+y;
    if(l==r){siz[x]+=siz[y];return x;}
    int mid=l+r>>1;
    son[x][0]=merge(son[x][0],son[y][0],l,mid);
    son[x][1]=merge(son[x][1],son[y][1],mid+1,r);
    siz[x]=siz[son[x][0]]+siz[son[x][1]];
    return x;
}
inline int query(int p,int l,int r,int v){
    if(l==r)return siz[p];
    int mid=l+r>>1;
    if(v<=mid)return query(son[p][0],l,mid,v)+siz[son[p][1]];
    return query(son[p][1],mid+1,r,v);
}
inline void dfs(int p){
    for(int i=first[p];i;i=e[i].next){
        int v=e[i].v;
        dfs(v);
        rt[p]=merge(rt[p],rt[v],1,sig);
    }
    ans[p]=query(rt[p],1,sig,a[p])-1;
}
int main(){
    n=read();
    for(int i=1;i<=n;++i)a[i]=b[i]=read();
    sort(b+1,b+n+1),sig=unique(b+1,b+n+1)-b-1;
    for(int i=1;i<=n;++i)modify(rt[i],1,sig,(a[i]=(lower_bound(b+1,b+sig+1,a[i])-b)));
    for(int i=2;i<=n;++i)add(read(),i);
    dfs(1);
    for(int i=1;i<=n;++i)write(ans[i]),puts("");
    return 0;
}

巴特西

2018.08.27 [Usaco2017 Jan]Promotion Counting（线段树合并）

描述

输入

输出

样例输入

样例输出

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