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NC51119. K-th Number

描述

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

输入描述

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer . 
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k and represents the question Q(i, j, k).

输出描述

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

示例1

输入:

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

输出:

5
6
3

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++(g++ 7.5.0) 解法, 执行用时: 176ms, 内存消耗: 4168K, 提交时间: 2023-01-20 18:25:21

#include<iostream>
using namespace std;

const int N=1e5+10,INF=1e9;
int n,m,t,c[N],ans[N];
struct rec{int op,x,y,z;}q[2*N],lq[2*N],rq[2*N];

int ask(int x)
{
    int toreturn=0;
    for(;x;x-=(x&-x)) toreturn+=c[x];
    return toreturn;
}

void add(int x,int y)
{
    for(;x<=n;x+=(x&-x)) c[x]+=y;
}

void solve(int lval,int rval,int st,int ed)
{
    if(st>ed) return;
    if(lval==rval)
    {
        for(int i=st;i<=ed;i++)
            if(q[i].op>0) ans[q[i].op]=lval;
        return;
    }
    int mid=(lval+rval)>>1;
    int lt=0,rt=0;
    for(int i=st;i<=ed;i++)
    {
        if(q[i].op==0)
        {
            if(q[i].y<=mid) add(q[i].x,1),lq[++lt]=q[i];
            else rq[++rt]=q[i];
        }
        else
        {
            int cnt=ask(q[i].y)-ask(q[i].x-1);
            if(cnt>=q[i].z) lq[++lt]=q[i];
            else q[i].z-=cnt,rq[++rt]=q[i]; 
        }
    }
    for(int i=ed;i>=st;i--)
        if(q[i].op==0&&q[i].y<=mid) add(q[i].x,-1);
    for(int i=1;i<=lt;i++) q[st+i-1]=lq[i];
    for(int i=1;i<=rt;i++) q[st+lt+i-1]=rq[i];
    solve(lval,mid,st,st+lt-1);
    solve(mid+1,rval,st+lt,ed); 
}

int main(void)
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        int temp;
        cin>>temp;
        q[++t].op=0,q[t].x=i,q[t].y=temp;
    }
    for(int i=1;i<=m;i++)
    {
        int l,r,k;
        cin>>l>>r>>k;
        q[++t].op=i,q[t].x=l,q[t].y=r,q[t].z=k;
    }
    solve(-INF,INF,1,t);
    for(int i=1;i<=m;i++) cout<<ans[i]<<endl;
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 93ms, 内存消耗: 38676K, 提交时间: 2020-09-09 13:25:17

#include<cstdio>

const int INF=1e9+1,M=1e5+10,N=50e5+10;
int v,op,t=0;
int d[M],tot[N],root[M],ls[N],rs[N];


void ins(int x,int &now,int l,int r)
{
	if(!now)now=++t;
	tot[now]=op+tot[x];
	int mid=(l+r)>>1;
	if(l==r)return ;
	if(v<=mid)
	{
		rs[now]=rs[x];
		ins(ls[x],ls[now],l,mid);
	}
	else
	{
		ls[now]=ls[x];
		ins(rs[x],rs[now],mid+1,r);
	}
}

int temp,l,r;

int findk(int x,int y,int k)
{
	int xx=root[x-1],yy=root[y];
	l=-INF;r=INF;
	while(l<r)
	{
		int mid=(l+r)>>1;//向下取整 
		temp=tot[ls[yy]]-tot[ls[xx]];
		if(k<=temp)
		{
			xx=ls[xx];yy=ls[yy];
			r=mid;
		}
		else
		{
			yy=rs[yy];xx=rs[xx];
			k-=temp;
			l=mid+1;
		}
	}
	return l;
}

int main()
{
	int n,m;scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&d[i]);
		v=d[i];op=1;
		ins(root[i-1],root[i],-INF,INF);
	}
	
	for(int i=1;i<=m;i++)
	{
		int x,y,c;scanf("%d %d %d",&x,&y,&c);
		printf("%d\n",findk(x,y,c));
	}
	return 0;
}

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