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

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

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]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

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

 

Sample Output

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[N],sorted[N],val[18][N],toleft[18][N];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[N],val[18][N];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;}