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NC24048. [USACO 2017 Jan P]Promotion Counting

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…N (), 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).

输入描述

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.

输出描述

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

示例1

输入:

5
804289384
846930887
681692778
714636916
957747794
1
1
2
3

输出:

2
0
1
0
0

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 50ms, 内存消耗: 6104K, 提交时间: 2019-08-09 15:13:35

#include<algorithm>
#include<cstdio>
#define Re register int
#define F(a,b) for(i=a;i<=b;++i)
const int N=1e5+3;
int x,i,n,t,C[N],Q[N],ip[N],ans[N],nex[N],head[N];
struct A{int x,i;bool operator<(A b)const{return x<b.x;};}a[N];
inline void add(Re x,Re y){Q[++t]=y,nex[t]=head[x],head[x]=t;}
inline void in(Re &x){
    x=0;char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
}
inline void addx(Re x){while(x<=n)++C[x],x+=x&(-x);}
inline int ask(Re x){int ans=0;while(x)ans+=C[x],x-=x&(-x);return ans;}
inline void dfs(Re x){
    ans[x]-=ask(n)-ask(ip[x]);
    for(Re i=head[x];i;i=nex[i])dfs(Q[i]);
    ans[x]+=ask(n)-ask(ip[x]);
    addx(ip[x]);
}
int main(){
    in(n);
    F(1,n)in(a[i].x),a[i].i=i;
    std::sort(a+1,a+n+1);
    F(1,n)ip[a[i].i]=i;
    F(2,n)in(x),add(x,i);
    dfs(1);
    F(1,n)printf("%d\n",ans[i]); 
}

C++11(clang++ 3.9) 解法, 执行用时: 87ms, 内存消耗: 6556K, 提交时间: 2019-07-01 21:30:36

#include<cstdio>
#include<algorithm>
#define F(i,a,b) for(int i=a;i<=b;++i)
#define eF(i,u) for(int i=h[u];i;i=nxt[i])
int n,q;
int a[100001],O[100001],Ans[100001],b[100001];
inline void I(int x){for(;x<=n;++b[x],x+=x&-x);}
inline int Q(int x){int sum=0;for(;x;sum+=b[x],x-=x&-x);return sum;}
int h[100001],nxt[100001],to[100001],tot;
inline void ins(int x,int y){nxt[++tot]=h[x];to[tot]=y;h[x]=tot;}
inline bool cmp(int p1,int p2){return a[p1]>a[p2];}
void dfs(int u){
    Ans[u]=-Q(a[u]);
    eF(i,u) dfs(to[i]);
    Ans[u]+=Q(a[u]);
    I(a[u]);
}
int main(){
    scanf("%d",&n);
    F(i,1,n) scanf("%d",a+i), O[i]=i;
    std::sort(O+1,O+n+1,cmp);
    F(i,1,n) a[O[i]]=i;
    int x;
    F(i,2,n) scanf("%d",&x), ins(x,i);
    dfs(1);
    F(i,1,n) printf("%d\n",Ans[i]);
    return 0;
}

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