POJ-2104-K-th Number(函数式线段树)
2014-07-28 10:30
381 查看
Description
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.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
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 (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
Sample Output
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
Source
Northeastern Europe 2004, Northern Subregion
思路:离散化,插入,查询。详见代码。
#include <cstdio>
#include <algorithm>
using namespace std;
struct S{
int val,id,pos;
}node[100005];
bool cmpval(struct S a,struct S b)
{
return a.val<b.val;
}
bool cmpid(struct S a,struct S b)
{
return a.id<b.id;
}
int T[100005],ls[4000000],rs[4000000],vv[100005],cnum[4000000],nodenum,cnt,n,m;
void insert(int s,int e,int pos,int pre,int &x)
{
x=++nodenum;
ls[x]=ls[pre];
rs[x]=rs[pre];
cnum[x]=cnum[pre]+1;
if(s!=e)
{
int mid=(s+e)>>1;
if(pos<=mid) insert(s,mid,pos,ls[pre],ls[x]);
else insert(mid+1,e,pos,rs[pre],rs[x]);
}
}
int query(int a,int b,int s,int e,int k)
{
if(s==e) return vv[s];
else
{
int mid=(s+e)>>1;
if(cnum[ls[b]]-cnum[ls[a]]>=k) return query(ls[a],ls[b],s,mid,k);//跟第b棵树的左子树和第a棵树的左子树的差比较
else return query(rs[a],rs[b],mid+1,e,k-cnum[ls[b]]+cnum[ls[a]]);
}
}
int main()
{
int i,a,b,k;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
scanf("%d",&node[i].val);
node[i].id=i;
vv[i]=node[i].val;
}
sort(vv+1,vv+n+1);
cnt=unique(vv+1,vv+n+1)-vv-1;//去重
sort(node+1,node+n+1,cmpval);
int p=1;
for(i=1;i<=n;i++)//离散化
{
if(node[i].val!=vv[p]) p++;
node[i].pos=p;
}
sort(node+1,node+n+1,cmpid);
T[0]=ls[0]=rs[0]=cnum[0]=nodenum=0;
for(i=1;i<=n;i++) insert(1,cnt,node[i].pos,T[i-1],T[i]);
while(m--)
{
scanf("%d%d%d",&a,&b,&k);
printf("%d\n",query(T[a-1],T[b],1,cnt,k));
}
}
}
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.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
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 (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3
Sample Output
5 6 3
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
Source
Northeastern Europe 2004, Northern Subregion
思路:离散化,插入,查询。详见代码。
#include <cstdio>
#include <algorithm>
using namespace std;
struct S{
int val,id,pos;
}node[100005];
bool cmpval(struct S a,struct S b)
{
return a.val<b.val;
}
bool cmpid(struct S a,struct S b)
{
return a.id<b.id;
}
int T[100005],ls[4000000],rs[4000000],vv[100005],cnum[4000000],nodenum,cnt,n,m;
void insert(int s,int e,int pos,int pre,int &x)
{
x=++nodenum;
ls[x]=ls[pre];
rs[x]=rs[pre];
cnum[x]=cnum[pre]+1;
if(s!=e)
{
int mid=(s+e)>>1;
if(pos<=mid) insert(s,mid,pos,ls[pre],ls[x]);
else insert(mid+1,e,pos,rs[pre],rs[x]);
}
}
int query(int a,int b,int s,int e,int k)
{
if(s==e) return vv[s];
else
{
int mid=(s+e)>>1;
if(cnum[ls[b]]-cnum[ls[a]]>=k) return query(ls[a],ls[b],s,mid,k);//跟第b棵树的左子树和第a棵树的左子树的差比较
else return query(rs[a],rs[b],mid+1,e,k-cnum[ls[b]]+cnum[ls[a]]);
}
}
int main()
{
int i,a,b,k;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
scanf("%d",&node[i].val);
node[i].id=i;
vv[i]=node[i].val;
}
sort(vv+1,vv+n+1);
cnt=unique(vv+1,vv+n+1)-vv-1;//去重
sort(node+1,node+n+1,cmpval);
int p=1;
for(i=1;i<=n;i++)//离散化
{
if(node[i].val!=vv[p]) p++;
node[i].pos=p;
}
sort(node+1,node+n+1,cmpid);
T[0]=ls[0]=rs[0]=cnum[0]=nodenum=0;
for(i=1;i<=n;i++) insert(1,cnt,node[i].pos,T[i-1],T[i]);
while(m--)
{
scanf("%d%d%d",&a,&b,&k);
printf("%d\n",query(T[a-1],T[b],1,cnt,k));
}
}
}
相关文章推荐
- POJ 2104 K-th Number (划分树 函数式线段树)
- poj-2104 K-th Number[主席树/函数式线段树/可持久化线段树]
- poj 2104 K-th Number(函数式线段树)
- [POJ 2104]K-th Number (不带修改的区间k大,函数式线段树)
- poj 2104 K-th Number 函数式线段树
- POJ2104 K-th number 函数式线段树
- POJ 2104 K-th number 主席树 函数式线段树
- poj2104 K-th Number (函数式线段树)
- [POJ]2104 K-th Number 主席树&线段树合并&整体二分
- poj 2104 K-th Number(线段树)
- 主席树(可持久化线段树)讲解 [POJ 2104] K-th Number
- POJ 2104 K-th Number 线段树
- poj 2104 K-th Number(可持久线段树)
- POJ 2104 K-th Number (可持久化线段树)
- POJ_2104_K-th Number_线段树(归并树)
- POJ-2104 K-th Number(线段树[归并树]-区间第k大)
- 【原】 POJ 2104 K-th Number 线段树 划分树 合并树 解题报告
- 用线段树求区间第K大(POJ 2104 K-th Number)
- POJ2104————K-th Number(线段树,二分法)
- poj 2104 K-th Number(划分线段树)