POJ2452---Sticks Problem(单调栈+RMQ,方法不够优秀)

Description

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by S1, S2, S3, …Sn. After measuring the length of each stick Sk (1 <= k <= n), she finds that for some sticks Si and Sj (1<= i < j <= n), each stick placed between Si and Sj is longer than Si but shorter than Sj.

Now given the length of S1, S2, S3, …Sn, you are required to find the maximum value j - i.

Input

The input contains multiple test cases. Each case contains two lines.

Line 1: a single integer n (n <= 50000), indicating the number of sticks.

Line 2: n different positive integers (not larger than 100000), indicating the length of each stick in order.

Output

Output the maximum value j - i in a single line. If there is no such i and j, just output -1.

Sample Input

4

5 4 3 6

4

6 5 4 3

Sample Output

1

-1

Source

我先用一个单调栈,找出每个值A[i]A[i]作为最大值时,最远往左可以扩展到哪里

然后rmq求出这个区间里的最小值A[j]A[j],那么这时的区间就是i−ji - j

感觉nlognnlogn还是不够优秀

/*************************************************************************
    > File Name: POJ2452.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月15日 星期五 17时12分22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 = 50010;
int arr
;
PLL Stack
;
int Top;
int L
;
int dp
[20];
int _hash[N << 1];
int LOG
;

void initRMQ(int n) {
    for (int i = 1; i <= n; ++i) {
        dp[i][0] = arr[i];
    }
    for (int j = 1; j <= LOG
; ++j) {
        for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
            dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
        }
    }
}

int ST(int l, int r) {
    int k = LOG[r - l + 1];
    return min(dp[l][k], dp[r - (1 << k) + 1][k]);
}

int main() {
    int n;
    LOG[0] = -1;
    for (int i = 1; i <= 50000; ++i) {
        LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1;
    }
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &arr[i]);
            _hash[arr[i]] = i;
        }
        initRMQ(n);
        int ans = -1;
        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) {
            int l = L[i];
            if (l == i) {
                continue;
            }
            int p = _hash[ST(l, i)];
            ans = max(ans, i - p);
        }
        printf("%d\n", ans);
    }
    return 0;
}