Codeforces Round #305 (Div. 1)B - Mike and Feet 线段树

B. 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 from1 ton from left to right.i-th bear is exactlya_i 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 strength of a group is the minimum height of the bear in that group.

Mike is a curious to know for each x such that1 ≤ x ≤ n the maximum strength among all groups of sizex.

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, ..., a_n (1 ≤ a_i ≤ 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

101 2 3 4 5 4 3 2 1 6

Output

6 4 4 3 3 2 2 1 1 1 

题意：对于长度为x的子序列，每个序列的score为这个序列中的最小值，输出长度为x 的子序列的最大值

思路：首先按照值排序，按照从小到大的顺序添加到set中，对于新添加的数，如果他在set中的两边的位置分别为l，r的话，那么他可以修改长度小于r-l-1的子序列的score，使它们变大，线段树维护长度为x的子序列的score的最大值

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=200010;int N;struct node{    int x,id;    bool operator<(const node &a)const    {        return x<a.x;    }}a[maxn];set<int> mp;set<int>::iterator it;int ans[maxn];struct IntervalTree{    int maxv[maxn<<2];    int setv[maxn<<2];    void build(int o,int l,int r)    {        maxv[o]=setv[o]=0;        if(l==r)return ;        int mid=(l+r)>>1;        build(o<<1,l,mid);        build(o<<1,mid+1,r);    }    void update(int o,int l,int r,int q1,int q2,int x)    {        if(q1<=l&&r<=q2)        {            setv[o]=x;            maxv[o]=x;            return ;        }        pushdown(o);        int mid=(l+r)>>1;        if(q1<=mid)update(o<<1,l,mid,q1,q2,x);        if(q2>mid)update(o<<1|1,mid+1,r,q1,q2,x);    }    void pushdown(int o)    {        if(setv[o])        {            setv[o<<1]=setv[o<<1|1]=setv[o];            maxv[o<<1]=maxv[o<<1|1]=setv[o];            setv[o]=0;        }    }    int query(int o,int l,int r,int x)    {        if(l==r)return maxv[o];        pushdown(o);        int mid=(l+r)>>1;        if(x<=mid)return query(o<<1,l,mid,x);        return query(o<<1|1,mid+1,r,x);    }}tree;int main(){    scanf("%d",&N);    for(int i=1;i<=N;i++)    {        scanf("%d",&a[i].x);        a[i].id=i;    }    sort(a+1,a+1+N);       mp.insert(0);mp.insert(N+1);    int l,r;    tree.build(1,1,N);    for(int i=1;i<=N;i++)    {        it=mp.lower_bound(a[i].id);        r=*it;it--;l=*it;        if(r-l-1>=1)tree.update(1,1,N,1,r-l-1,a[i].x);        mp.insert(a[i].id);    }    for(int i=1;i<=N;i++)        ans[i]=tree.query(1,1,N,i);    for(int i=1;i<=N;i++)        printf("%d ",ans[i]);    printf("\n");    return 0;}

还可以用更简单的方法（下面的代码转自codeforces）：

#include <bits/stdc++.h>using namespace std;template <typename T> bool maxup(T &a, const T &b) { return a < b ? (a = b, true) : false; }template <typename T> bool minup(T &a, const T &b) { return a > b ? (a = b, true) : false; }const int MAX = 200100;int n;int a[MAX], low[MAX], high[MAX], m[MAX];int main() {    scanf("%d", &n);    for (int i = 1; i <= n; i++)        scanf("%d", &a[i]);    stack<int> up;    up.push(0);    for (int i = 1; i <= n; i++) {        while (a[i] <= a[up.top()])            up.pop();        low[i] = up.top();        up.push(i);    }    up = stack<int>();    up.push(n + 1);    for (int i = n; i > 0; i--) {        while (a[i] <= a[up.top()])            up.pop();        high[i] = up.top();        up.push(i);    }    for (int i = 1; i <= n; i++)        maxup(m[high[i] - low[i] - 1], a[i]);    for (int i = n - 1; i >= 1; i--)        maxup(m[i], m[i + 1]);    for (int i = 1; i <= n; i++)        printf("%d ", m[i]);}