[HDU 2665]Kth number
[HDU 2665]Kth number
题目
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
OUTPUT
2
解题报告
初识主席树
虽然只会静态的2333
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 //#define mem(x) memset((x),0,sizeof(x)) 7 inline int read(){ 8 int sum(0); 9 char ch(getchar()); 10 for(;ch<'0'||ch>'9';ch=getchar()); 11 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 12 return sum; 13 } 14 int T; 15 int n,m; 16 int a[100005],num[100005]; 17 int cnt,size; 18 int rt[2000005],sum[2000005],lch[2000005],rch[2000005]; 19 inline void clear(){ 20 // mem(a),mem(num),mem(rt),mem(sum),mem(lch),mem(rch); 21 cnt=size=n=m=0; 22 } 23 inline void build(int &x,int l,int r){ 24 x=++cnt; 25 sum[x]=0; 26 if(l==r){ 27 lch[x]=rch[x]=0; 28 return; 29 } 30 int mid((l+r)>>1); 31 build(lch[x],l,mid); 32 build(rch[x],mid+1,r); 33 } 34 inline void update(int &x,int las,int pos,int l,int r){ 35 x=++cnt; 36 lch[x]=lch[las]; 37 rch[x]=rch[las]; 38 sum[x]=sum[las]+1; 39 if(l==r) 40 return; 41 int mid((l+r)>>1); 42 if(pos<=mid) 43 update(lch[x],lch[las],pos,l,mid); 44 else 45 update(rch[x],rch[las],pos,mid+1,r); 46 } 47 inline int query(int ll,int rr,int l,int r,int k){ 48 if(l==r) 49 return l; 50 int mid((l+r)>>1); 51 int cnt(sum[lch[rr]]-sum[lch[ll]]); 52 if(k<=cnt) 53 return query(lch[ll],lch[rr],l,mid,k); 54 return query(rch[ll],rch[rr],mid+1,r,k-cnt); 55 } 56 int main(){ 57 T=read(); 58 while(T--){ 59 clear(); 60 n=read(),m=read(); 61 for(int i=1;i<=n;++i) 62 num[i]=a[i]=read(); 63 sort(num+1,num+n+1); 64 size=unique(num+1,num+n+1)-num-1; 65 for(int i=1;i<=n;++i) 66 a[i]=lower_bound(num+1,num+size+1,a[i])-num; 67 build(rt[0],1,size); 68 for(int i=1;i<=n;++i) 69 update(rt[i],rt[i-1],a[i],1,size); 70 for(int i=1;i<=m;++i){ 71 int l(read()),r(read()),k(read()); 72 int ans(query(rt[l-1],rt[r],1,size,k)); 73 printf("%d\n",num[ans]); 74 } 75 } 76 }
