HDU-2665-Kth number(划分树)

[align=left]Problem Description[/align]
Give you a sequence and ask you the kth big number of a inteval.

[align=left]Input[/align]
The first line is the number of the test cases.

For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.

The second line contains n integers, describe the sequence.

Each of following m lines contains three integers s, t, k.

[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

[align=left]Output[/align]
For each test case, output m lines. Each line contains the kth big number.

[align=left]Sample Input[/align]

1
10 1
1 4 2 3 5 6 7 8 9 0
1 3 2

[align=left]Sample Output[/align]

2

[align=left]Source[/align]
HDU男生专场公开赛——赶在女生之前先过节（From
WHU）

思路：划分树板子题。

#include <cstdio>
#include <algorithm>
using namespace std;

int n,num[100005],sorted[100005],node[20][100005],sum[20][100005];

void build(int c,int s,int e)//第c层，s到e
{
int mid,lm,lp,rp,i;

mid=(s+e)>>1;
lm=0;
lp=s;//左子树的头指针
rp=mid+1;//右子树的头指针

for(i=s;i<=mid;i++) if(sorted[i]==sorted[mid]) lm++;//求出s到e有多少个数等于sorted[mid]

for(i=s;i<=e;i++)
{
if(i==s) sum[c][i]=0;//计算出s到i之间有多少个被分到了左子树
else sum[c][i]=sum[c][i-1];

if(node[c][i]==sorted[mid])
{
if(lm)
{
lm--;
sum[c][i]++;
node[c+1][lp++]=node[c][i];
}
else node[c+1][rp++]=node[c][i];
}
else if(node[c][i]<sorted[mid])
{
sum[c][i]++;
node[c+1][lp++]=node[c][i];
}
else node[c+1][rp++]=node[c][i];
}

if(s!=e)
{
build(c+1,s,mid);
build(c+1,mid+1,e);
}
}

int query(int c,int s,int e,int l,int r,int k)
{
if(s==e) return node[c][s];
else
{
int ls,rs,mid;

mid=(s+e)>>1;

if(s==l)//特判
{
ls=0;
rs=sum[c][r];
}
else
{
ls=sum[c][l-1];
rs=sum[c][r]-ls;
}

if(rs>=k) return query(c+1,s,mid,s+ls,s+ls+rs-1,k);//要查询的数在左子树
else return query(c+1,mid+1,e,mid-s+1+l-ls,mid-s+1+r-ls-rs,k-rs);//要查询的数在右子树
}
}

int main()
{
int T,m,i,a,b,k;

scanf("%d",&T);

while(T--)
{
scanf("%d%d",&n,&m);

for(i=1;i<=n;i++)
{
scanf("%d",&num[i]);

node[0][i]=sorted[i]=num[i];
}

sort(sorted+1,sorted+n+1);

build(0,1,n);

while(m--)
{
scanf("%d%d%d",&a,&b,&k);

printf("%d\n",query(0,1,n,a,b,k));
}
}
}