码迷,mamicode.com
首页 > 其他好文 > 详细

Codeforces Round #305 (Div. 2)D---Mike and Feet(单调栈)

时间:2015-05-27 22:58:31      阅读:178      评论:0      收藏:0      [点我收藏+]

标签:单调栈

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 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

单调栈处理出以每个数为最小值,可以得到的最长区间
然后排个序

/*************************************************************************
    > File Name: D.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月27日 星期三 15时45分40秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

static const int N = 200010;
int L[N], R[N];
PLL Stack[N];
int Top;
int arr[N];
PLL use[N];

int cmp(PLL a, PLL b) {
    return a.first > b.first;
}

int main() {
    int n;
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &arr[i]);
            use[i] = make_pair(arr[i], i);
        }
        Top = 0;
        for (int i = n; i >= 1; --i) {
            if (!Top) {
                Stack[++Top] = make_pair(arr[i], i);
            }
            else {
                while (Top) {
                    PLL u = Stack[Top];
                    if (u.first <= arr[i]) {
                        break;
                    }
                    --Top;
                    L[u.second] = i + 1;
                }
                Stack[++Top] = make_pair(arr[i], i);
            }
        }
        while (Top) {
            PLL u = Stack[Top];
            --Top;
            L[u.second] = 1;
        }
        for (int i = 1; i <= n; ++i) {
            if (!Top) {
                Stack[++Top] = make_pair(arr[i], i);
            }
            else {
                while (Top) {
                    PLL u = Stack[Top];
                    if (u.first <= arr[i]) {
                        break;
                    }
                    --Top;
                    R[u.second] = i - 1;
                }
                Stack[++Top] = make_pair(arr[i], i);
            }
        }
        while (Top) {
            PLL u = Stack[Top];
            --Top;
            R[u.second] = n;
        }
        int len = 1;
        sort(use + 1, use + 1 + n, cmp);
        for (int i = 1; i <= n; ++i) {
            int l = L[use[i].second];
            int r = R[use[i].second];
            for (int j = len; j <= r - l + 1; ++j) {
                printf("%d ", use[i].first);
                ++len;
            }
            if (len > n) {
                break;
            }
        }
        printf("\n");
    }
    return 0;
}

Codeforces Round #305 (Div. 2)D---Mike and Feet(单调栈)

标签:单调栈

原文地址:http://blog.csdn.net/guard_mine/article/details/46051419

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有 京ICP备13008772号-2
迷上了代码!