LowerBound
Time Limit: 1 Sec Memory Limit: 128 MB
Submissions: 8 Solved: 7
Description
You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 2*10^9, 1 ≤ N ≤ 100000 ). A query is defined as follows:
(L,R,V) : find the smallest number of A[i] such that L<=i<=R and A[i]>V, if not exist, output
“not exist”.
Given M queries, your program must output the results of these queries.
Input
The first line contains a single integer T, the number of test cases.
For each case, there are two integers N and M (1<= N, M <=100000).
The next line contain N elements.
A1 A2 … AN
The next M lines contain the operation in following form.
L1 R1 V1
L2 R2 V2
…
LM RM VM
Output
For each question, output the answer in one line.
Sample Input
1 5 5 5 4 3 2 1 1 5 2 2 4 0 3 5 3 1 4 3 2 5 1
Sample Output
3 2 not exist 4 2
HINT
Source
The 8th(2013) ACM Programming Contest of HUST
在划分树基础上进行二分
n*log(n)*log(n)
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn = 110000;
int rec[maxn];
struct node{
int left;
int right;
int mid;
}tree[maxn*5];
struct point{
int value;
int right;
bool is_right;
}po[20][maxn];
int n,m;
int build_tree(int left,int right,int deepth,int pos){
tree[pos].left=left;
tree[pos].right=right;
tree[pos].mid=(left+right)>>1;
int mid=rec[tree[pos].mid],l=left,r=tree[pos].mid+1,i,j,k;
point *now=po[deepth-1],*then=po[deepth];
int ll=0;
for(i=tree[pos].mid;i>=left;i--){
if(rec[i]==mid) ll++;
}
for(i=left;i<=right;i++){
if(now[i].value==mid && ll){
ll--;
then[l++].value=mid;
now[i].is_right=false;
continue;
}
if(now[i].value>=mid){
then[r++].value=now[i].value;
now[i].is_right=true;
}else{
then[l++].value=now[i].value;
now[i].is_right=false;
}
}
now[left].right=now[left].is_right;
for(i=left+1;i<=right;i++){
now[i].right=now[i-1].right+now[i].is_right;
}
if(left == right) return 0;
build_tree(left,tree[pos].mid,deepth+1,pos<<1);
build_tree(tree[pos].mid+1,right,deepth+1,(pos<<1)+1);
return 0;
}
int query(int left,int right,int k,int deepth,int pos){
int num=0;
point *now=po[deepth];
if(tree[pos].left == tree[pos].right) return now[tree[pos].left].value;
num=right-left+1-(now[right].right-now[left].right+now[left].is_right);
if(num>=k)
return query(left-now[left].right+now[left].is_right,right-now[right].right,k,deepth+1,pos<<1);
else
return query(tree[pos].mid+now[left].right+1-now[left].is_right,\
tree[pos].mid+now[right].right,k-num,deepth+1,(pos<<1)+1);
}
int find_ans(int a,int b,int num){
int mid,l=1,r=b-a+1,now,ans=-2000000010;
while(l<=r){
mid=(l+r)>>1;
now=query(a,b,mid,0,1);
if(now <= num) l=mid+1;
else{
r=mid-1;
ans=now;
}
}
return ans;
}
int main(){
int i,j,a,b,k,t,ans;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++){
scanf("%d",&rec[i]);
po[0][i].value=rec[i];
}
sort(rec+1,rec+1+n);
build_tree(1,n,1,1);
while(m--){
scanf("%d%d%d",&a,&b,&k);
ans=find_ans(a,b,k);
if(ans>k){
printf("%d\n",ans);
}else{
printf("not exist\n");
}
}
}
return 0;
}