Codeforces Round #305 B题思维+贪心

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最新推荐文章于 2024-12-01 03:34:23 发布

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本博客讨论了在给定熊群高度序列的情况下，如何计算不同子区间长度的最大强度值。通过分析熊群序列，实现了一种高效的算法来解决这一问题。

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B. Mike and Feet
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 strength of 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 × 10^5), the number of bears.

The second line contains n integers separated by space, a1, a2, …, an (1 ≤ ai ≤ 10^9), 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

题意：x从1到n，求长度x的这些子区间的最大值，每个区间的最大值定义为该区间的最小值

解法：

求出每个值x向左和右以x为最小值所能延伸的范围width
然后贪心求解，对于位置site，它有两个属性，height和width，以height为第一优先排序
每次从排序后的集合取最优先的值，如果他的width大与当前取的次数，那么输出这个height，否则重复当前步骤

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cassert>
using namespace std ;

const int N = 2e5 + 11 ;

struct Node {
    int h , w ;
};

int tol[N] , tor[N] ;
int height[N] ;
Node arr[N] ;

bool cmp(const Node& a , const Node& b) {
    if(a.h == b.h) return a.w > b.w ;
    return a.h > b.h ;
}


int main() {
    //freopen("data.in", "r" , stdin);
    int n ;
    while(scanf("%d" ,&n)==1) {
        for(int i = 1 ; i <= n;  ++i) scanf("%d" ,&height[i]) ;
        tol[1] = 1 ; tor[n] = n ;
        for(int i = 1 ; i <= n ; ++i) {
            int tmp = i ;
            while(tmp > 1 && height[tmp-1] >= height[i]) tmp = tol[tmp-1] ;
            tol[i] = tmp ;
        }
        for(int i = n-1 ; i >= 1 ; --i) {
            int tmp = i ;
            while(tmp < n && height[tmp+1] >= height[i]) tmp = tor[tmp+1] ;
            tor[i] = tmp ;
        }
        for(int i = 1 ; i <= n ; ++i) {
            arr[i-1].w = (tor[i]-tol[i]+1) ;
            arr[i-1].h = height[i] ;
        }
        sort(arr , arr+n , cmp) ;
        int last = 0 ;
        for(int i = 0 ; i < n ; ++i) {
            if(arr[i].w > last) {
                printf("%d" , arr[i].h) ;
                ++last ;
                if(last == n) {printf("\n") ;break;}
                else printf(" ") ;
                --i ;
            }
        }
    }
}