单调栈Minimum Sum

Posted 2022-12-10 biu~跃哥冲冲冲

Minimum Sum

题目描述

One day, Snuke was given a permutation of length $N,a_1,a_2,\dots ,a_N$, from his friend.
Find the following:

Constraints
$1\le N\le 200,000$
($a_1,a_2,\dots ,a_N$) is a permutation of (1,2,…,N).

输入

The input is given from Standard Input in the following format:
$N$
$a_1{a}_2\dots a_N$

输出

Print the answer.
Note that the answer may not fit into a 32-bit integer.

样例输入:

3
2 1 3

样例输出:

9

题目大意：其实就是计算给的那个式子，哈哈哈，为了方便起见，我拿题目所给样例模拟一下；

清晰的记得这道题当时用线段树给超时了，结果赛后搜博客加上和同学讨论许久我也还是不懂qwq,看到标记是单调栈，没整过呀，怎么搞，搜博客搜题解，看完后依然一脸懵逼，然后，就把他放那不管了，今天一个机遇又看了看单调栈，也不是太难理解，随后就又翻出这道题，写上一番提交后就A了，哈哈哈，还是很NICE的嘛！！！

解题思路：用单调栈找出每一个数的左边界和右边界，要想用到这个数，前提是它在这个区间内是最小的那个，不然的话就另有其数了，然后一个关键就是，怎样计算它的权重值：假如这个数是a[ i ],他的左边界为L,右边界为R,这里的L和R都是下标的表示,val=a[ i ]( i - L[ i ] )(R[ i ] - i )，据说是乘法原理，但这里我还是没整太明白，友友如若知晓的话可以联系博主，在此表示感谢啦！

AC代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
ll L[201010], R[201010], a[201010];
struct number 
    ll num, id;
st[201010];
inline ll read()
    ll  x = 0;
    bool f = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)    f |= ch == '-', ch = getchar();
    while ('0' <= ch && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;

int main()

    
    ll n, i;
    n = read();
    for (i = 1;i <= n;i++)
    
        a[i] = read();
        L[i] = 0;
        R[i] = n + 1;
    
    ll s = 0;
    for (i = 1;i <= n;i++)
    
        while (s && st[s].num >= a[i]) s--;
        if (s) L[i] = st[s].id;
        s++;
        st[s].num = a[i];
        st[s].id = i;
    
    s = 0;
    for (i = n;i >= 1;i--)
    
        while (s && st[s].num >= a[i]) s--;
        if (s) R[i] = st[s].id;
        s++;
        st[s].num = a[i];
        st[s].id = i;
    
    ll ans = 0;
    for (i = 1;i <= n;i++)
    
        ans += a[i] * (i - L[i]) * (R[i] - i);
    
    cout << ans << endl;
    return 0;

/*

2 1 3

*/