Codeforces Round #305 (Div. 2) D题 （线段树+RMQ）

D. Mike and Feet

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.



A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strengthof a group is the minimum height of the bear in that group.

Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.

Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.

The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.

Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

Sample test(s)

input

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1

先用ST算法处理出一段区间内最小值。二分查询出对于第i个数，以它为最小向左或向右可达的最值，得到区间长度为r-l+1。那么，对于group长度为1，2，3，....r-l+1最小值为第i个数的值。则区间更新这些组大小，取最大值。此处可以使用线段树维护。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=200050;
const int inf=1e9+107;

int feet
;
int st
[20];
int seg[N<<2];
int ans
;
void pushdown(int rt){
seg[rt<<1]=max(seg[rt<<1],seg[rt]);
seg[rt<<1|1]=max(seg[rt<<1|1],seg[rt]);
seg[rt]=-1;
}

void ST(int num){
for(int i=1;i<=num;i++)
st[i][0]=feet[i];
for(int j=1;j<20;j++)
for(int i=1;i<=num;i++){
if(i+(1<<j)-1 <= num){
st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);
}
}
}

void update(int rt,int l,int r,int L,int R,int val){
if(l<=L&&R<=r){
if(val>seg[rt]) seg[rt]=val; return;
}
int m=(L+R)>>1;
pushdown(rt);
if(r<=m){
update(rt<<1,l,r,L,m,val);
}
else if(l>=m+1){
update(rt<<1|1,l,r,m+1,R,val);
}
else{
update(rt<<1,l,r,L,m,val);
update(rt<<1|1,l,r,m+1,R,val);
}
}

void dfs(int rt,int L,int R){
if(L==R){
ans[L]=seg[rt];
return ;
}
pushdown(rt);
int m=(L+R)>>1;
dfs(rt<<1,L,m);
dfs(rt<<1|1,m+1,R);
}

int query(int l,int r){
int k=(int)((log(r-l+1))/log(2.0));
int maxl=min(st[l][k],st[r-(1<<k)+1][k]);
return maxl;
}

int main(){
int n;
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=(n<<2)+10;i++){
seg[i]=-1;
}
for(int i=1;i<=n;i++){
scanf("%d",&feet[i]);
}
ST(n);
for(int i=1;i<=n;i++){
//left
int l=1,r=i;
int L=i,R=i;
while(l<=r){
int m=(l+r)>>1;
int index=query(m,i);
if(index>=feet[i]){
r=m-1;
L=m;
}
else {
l=m+1;
}
}
//right
l=i,r=n;
while(l<=r){
int m=(l+r)>>1;
int index=query(i,m);
if(index>=feet[i]){
l=m+1;
R=m;
}
else {
r=m-1;
}
}
update(1,1,R-L+1,1,n,feet[i]);
}
dfs(1,1,n);
printf("%d",ans[1]);
for(int i=2;i<=n;i++)
printf(" %d",ans[i]);
printf("\n");
}
return 0;
}