CodeForces 548D （单调栈+伪DP）

最新推荐文章于 2021-09-10 19:15:13 发布

MrBlankIsAwesome

最新推荐文章于 2021-09-10 19:15:13 发布

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分类专栏： DP

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本文介绍了一道有趣的算法题——“熊群挑战”，利用单调栈求解不同长度连续子序列的最大最小值。通过对单调栈的巧妙应用，实现了高效地求解任意长度子序列的最大价值。

Mike and Feet
Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u
Submit

Status

Description
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 × 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 Input
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Source
Codeforces Round #305 (Div. 2)

给出一个数列，定义某个长度子序列（连续）的价值为其中最小的那个数，请输出每种长度序列的最大价值。
百度了一下是单调栈，但是感觉很像DP啊。。解析的话看注释

#include "cstring"
#include "cstdio"
#include "iostream"
#include "string.h"
#include "stack"
#include "cmath"
using namespace std;
int bear[200005];
int ans[200005];

typedef struct node
{
    int num,width;
}node;
stack<node> s;
void ins(int num,int width)
{
    node temp;
    temp.num=num;
    temp.width=width;
    s.push(temp);
    return;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        while(!s.empty())
            s.pop();
        memset(ans,0,sizeof(ans));
        memset(bear,0,sizeof(bear));
        for(int i=1;i<=n;i++)
            scanf("%d",&bear[i]);
        //通过单调栈维护出某个数的最长区间 不断更新某长度区间最优值
        //要往后多跑一位，因为如果要新全序列的值需要从n+1往回迭代
        for(int i=1;i<=n+1;i++)
        {
            int len=0;
            while(!s.empty())
            {
                node temp=s.top();
                //如果新插入的数比栈顶大，就不影响当前的连续区间
                if(temp.num<bear[i])
                    break;
                //如果新插入的数比栈顶小，那么就可以将当前连续区间压缩，同时更新各个长度的区间的值
                len+=temp.width;
                ans[len]=max(ans[len],temp.num);
                s.pop();
            }
            ins(bear[i],len+1);
        }
        for(int i=n-1;i>=1;i--)
        {
            ans[i]=max(ans[i+1],ans[i]);
        }
        printf("%d",ans[1]);
        for(int i=2;i<=n;i++)
            printf("% d",ans[i]);
        printf("\n");

    }
}