B. Knights of a Polygonal Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than kk other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim’s coins.

Now each knight ponders: how many coins he can have if only he kills other knights?

You should answer this question for each knight.

Input

The first line contains two integers nn and kk (1≤n≤105,0≤k≤min(n−1,10))(1≤n≤105,0≤k≤min(n−1,10)) — the number of knights and the number kk from the statement.

The second line contains nn integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤109)(1≤pi≤109) — powers of the knights. All pipi are distinct.

The third line contains nn integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤109)(0≤ci≤109) — the number of coins each knight has.

Output

Print nn integers — the maximum number of coins each knight can have it only he kills other knights.

题解：
按照a的值升序排列，在i前面的a都小于ai,所以去1~i-1 这段区间中前k大的k个数。
因为k<=10， 因此可以用线段树来维护区间前k大。
代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e5+7;
struct node
{
    int a,b,id;
}t[maxn];
struct pp
{
    ll ans;int id;
}s[maxn];
vector<int>G[maxn<<2];
int n,k,w[maxn],tol;
bool cmp(const node &x,const node &y)
{
    return x.a<y.a;
}
bool cmp1(const pp &x,const pp &y)
{
    return x.id<y.id;
}
void push_up(int rt)
{
    int L1=G[rt<<1].size(),L2=G[rt<<1|1].size(),l1=0,l2=0;
    while(G[rt].size()<k)
    {
        if(l1<L1&&l2<L2)
        {
            if(G[rt<<1][l1]>G[rt<<1|1][l2])
                G[rt].push_back(G[rt<<1][l1++]);
            else
                G[rt].push_back(G[rt<<1|1][l2++]);
        }
        else if(l1<L1)
        {
            G[rt].push_back(G[rt<<1][l1++]);
        }
        else if(l2<L2)
        {
            G[rt].push_back(G[rt<<1|1][l2++]);
        }
        else break;
    }
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        G[rt].push_back(t[l].b);
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    push_up(rt);
}
void query(int L,int R,int l,int r,int rt)
{
    if(l>=L&&r<=R)
    {
        for(int i=0;i<G[rt].size();i++)
            w[tol++]=G[rt][i];
        return;
    }
    int mid=(l+r)>>1;
    if(L<=mid)query(L,R,l,mid,rt<<1);
    if(R>mid)query(L,R,mid+1,r,rt<<1|1);
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&t[i].a);
        t[i].id=i;
    }
    for(int i=1;i<=n;i++)
        scanf("%d",&t[i].b);
    sort(t+1,t+n+1,cmp);
    build(1,n,1);
    s[1].ans=t[1].b;s[1].id=t[1].id;
    for(int i=2;i<=n;i++)
    {
        tol=0;
        s[i].ans=t[i].b;
        query(1,i-1,1,n,1);
        sort(w,w+tol);
        for(int j=tol-1,h=0;h<k&&j>=0;j--,h++)
            s[i].ans+=w[j];
        s[i].id=t[i].id;
    }
    sort(s+1,s+n+1,cmp1);
    for(int i=1;i<=n;i++)
        printf("%lld ",s[i].ans);
    return 0;
}