codeforces 5E Bindian Signalizing

E. Bindian Signalizing

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Everyone knows that long ago on the territory of present-day Berland there lived Bindian tribes. Their capital was surrounded by n hills, forming a circle. On each hill there was a watchman, who watched the neighbourhood day and night.

In case of any danger the watchman could make a fire on the hill. One watchman could see the signal of another watchman, if on the circle arc connecting the two hills there was no hill higher than any of the two. As for any two hills there are two different circle arcs connecting them, the signal was seen if the above mentioned condition was satisfied on at least one of the arcs. For example, for any two neighbouring watchmen it is true that the signal of one will be seen by the other.

An important characteristics of this watch system was the amount of pairs of watchmen able to see each other's signals. You are to find this amount by the given heights of the hills.

Input

The first line of the input data contains an integer number n (3 ≤ n ≤ 106), n — the amount of hills around the capital. The second line contains n numbers — heights of the hills in clockwise order. All height numbers are integer and lie between 1 and 109.

Output

Print the required amount of pairs.

题目的意思是有n座围成一个圈的山，站在两个山头能互相看见当且仅当：两座山之间没有比他们更高的山。

给出n座山的高度，输出能相互看见的pair数。

首先将输入的数字顺序预处理一下，找到最高的山峰，将这个圈从该处切断成链。

最简单的就是暴力了，对于一座山，枚举的判断其他山能否和他互相看见，这样的复杂度是O(n^2)，这题好像数据量略大，这样的复杂度是过不了的。

经过查看别人的解题报告，有这么一种dp的方法：

假设对于i，我们已经计算出了left[i], right[i], same[i]，其中left[i]表示i左侧比第一个比i高的位置，right[i]表示i右侧第一个比i高的位置，same[i]表示从i到right[i]的左开右闭区间内高度等于i的山的数目。

简而言之，left和right是位置，而same是数目。

那么对于一座山i而言，它可以和left[i] 和 right[i]都构成能互相看见的pair，并且和i到right[i]之间所有高度等于i的山构成能互相看见的pair。

所以问题就是计算left数组、right数组和same数组。

可以考虑用动态规划的思想来做，具体的见代码吧，但是这样的一个求left数组的算法的复杂度不是很好分析，我感觉和kmp算法的复杂度有点类似，好难分析。。。。。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <stack>#include <map>#include <queue>#include <algorithm>using namespace std;int main() {    int n;    scanf("%d", &n);    vector<int> height(n);    for (int i = 0; i < n; i++) scanf("%d", &(height[i]));    vector<int>::iterator highest = max_element(height.begin(), height.end());    //cut the circle, make a chain    vector<int> num(highest, height.end());    num.reserve(n);    copy(height.begin(), highest, num.end());    //left[i] is the nearest bigger number in the left of num[i]    vector<int> left(n, -1);    for (int i = 1; i < n; i++) {        int p = i - 1;        while (num[i] > num[p]) p = left[p];        left[i] = num[i] == num[p] ? left[p] : p;    }    //right[i] is the nearest bigger number in the right of num[i]    //same[i] is the number of the number equal to num[i] in [i, right[i]]    vector<int> right(n, n);    vector<int> same(n, 0);    for (int i = n - 2; i >= 0; i--) {        int p = i + 1;        while (p < n && num[i] > num[p]) p = right[p];        if (p >= n) {            right[i] = p;            continue;        }        right[i] = num[i] == num[p] ? right[p] : p;        same[i] = num[i] == num[p] ? same[p] + 1 : 0;    }    long long ans = 0;    for (int i = 0; i < n; i++) {        if (left[i] >= 0) ans += 1;        if (right[i] < n) ans += 1;        if (right[i] >= n && left[i] > 0) ans += 1;        ans += same[i];    }    printf("%I64d\n", ans);    return 0;}