hdu 2665 Kth number

Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3329 Accepted Submission(s): 1107



[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

分析：询问区间[s,t]内第k大的数

划分树：

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100001;
template<class T>
inline bool scan_d(T &ret){
char c;bool sgn;
if(c=getchar(),c==EOF)return 0;
while(c!='-'&&(c<'0'||c>'9'))c=getchar();
sgn=(c=='-');
ret=sgn?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9')ret=ret*10+(c-'0');
if(sgn)ret=-ret;
return 1;
}
int arr
,sorted
,val[18]
,toleft[18]
;
void build(int l,int r,int d){
if(l==r)return;
int m=(l+r)>>1;
int same=m-l+1,i=l,j=m+1,k;
for(k=l;k<=r;k++)if(val[d][k]<sorted[m])same--;
for(k=l;k<=r;k++){
if(val[d][k]<sorted[m])val[d+1][i++]=val[d][k];
else if(val[d][k]==sorted[m]&&same>0){
val[d+1][i++]=val[d][k];
same--;
}else val[d+1][j++]=val[d][k];
toleft[d][k]=toleft[d][l-1]+i-l;
}
build(l,m,d+1);
build(m+1,r,d+1);
}
int query(int L,int R,int l,int r,int d,int k){
if(l==r)return val[d][l];
int m=(L+R)>>1;
int cnt=toleft[d][r]-toleft[d][l-1];
if(cnt>=k){
int nl=L+toleft[d][l-1]-toleft[d][L-1];
int nr=nl+cnt-1;
return query(L,m,nl,nr,d+1,k);
}else{
int nr=r+toleft[d][R]-toleft[d][r];
int nl=nr-(r-l-cnt);
return query(m+1,R,nl,nr,d+1,k-cnt);
}
}
int main(){
int T,n,m,s,t,k,i;
scanf("%d",&T);
while(T--){
scan_d(n),scan_d(m);
for(i=1;i<=n;i++){
scan_d(val[0][i]);
sorted[i]=val[0][i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0);
while(m--){
scan_d(s),scan_d(t),scan_d(k);
printf("%d\n",query(1,n,s,t,0,k));
}
}
return 0;
}

归并树（但超时）

#include<cstdio>
const int N=100002;
struct MergeTree{
int l,r,m;
}a[N<<2];
int arr
,val[18]
;
void build(int l,int r,int d,int rt){
a[rt].l=l,a[rt].r=r,a[rt].m=(l+r)>>1;
if(l==r){val[d][l]=arr[l];return ;}
int m=a[rt].m;
build(l,m,d+1,rt<<1);
build(m+1,r,d+1,rt<<1|1);
int i=l,j=m+1,k=l;
while(i<=m&&j<=r){
if(val[d+1][i]<val[d+1][j])val[d][k++]=val[d+1][i++];
else val[d][k++]=val[d+1][j++];
}
while(i<=m)val[d][k++]=val[d+1][i++];
while(j<=r)val[d][k++]=val[d+1][j++];

}
int L,R;
int lowb(int key,int d,int rt){
if(a[rt].l==a[rt].r){
return key<=val[d][a[rt].m]?a[rt].m:a[rt].m+1;
}
if(key<=val[d][a[rt].m])return lowb(key,d,rt<<1);
else return lowb(key,d,rt<<1|1);
}
int find(int key,int d,int rt){
if(L<=a[rt].l&&R>=a[rt].r)return lowb(key,d,rt)-a[rt].l;
int ans=0;
if(L<=a[rt].m)ans+=find(key,d+1,rt<<1);
if(R>a[rt].m)ans+=find(key,d+1,rt<<1|1);
return ans;
}
int query(int l,int r,int k){
int m,tmp;
while(l<r){
m=(l+r+1)>>1;
tmp=find(val[0][m],0,1);
if(tmp<k)l=m;
else r=m-1;
}
return l;
}
int main(){
int T,n,m,i,k;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%d",arr+i);
build(1,n,0,1);
while(m--){
scanf("%d%d%d",&L,&R,&k);
printf("%d\n",val[0][query(1,n,k)]);
}
}
return 0;
}