TZOJ 4244 Sum(单调栈区间极差)

描述

Given a sequence, we define the seqence's value equals the difference between the largest element and the smallest element in the sequence. As an example, the value of sequence (3, 1, 7, 2) = 7-1 = 6.

Now, given a sequence S, output the sum of all value of consecutive subsequences.

输入

The first line has an integer N (2 ≤ N ≤ 300000) indicating the number of elements of the sequence. Then follows N lines, each line has a positive integer no larger than 10⁸ indicating an element of the sequence.

输出

Output the requested sum.

样例输入

4

3

1

7

2

样例输出

31

提示

The consecutive subsequence of the sequence (3, 1, 7, 2) has:
(3, 1), (1, 7), (7, 2), (3, 1, 7), (1, 7, 2), (3, 1, 7, 2)

The sum of all values equals 2+6+5+6+6+6=31

题意

求所有区间极差和

题解

N很大，考虑单调栈维护max和min，那么就是求ans=Σmax-Σmin

sum1代表最大值前缀和

sum2代表最小值前缀和

max单调栈，d[i]代表下标，b[i]代表值，tail1代表栈顶

min单调栈，e[i]代表下标，c[i]代表值，tail2代表栈顶

考虑加入一个a的贡献，sum1+=(d[tail1]-d[tail1-1])*a-弹出的Σ(d[tail1]-d[tail1-1])*b[tail1]，sum2+=(e[tail2]-e[tail2-1])*a-弹出的Σ(e[tail2]-e[tail2-1])*c[tail2]，就是计算后的sum1-sum2

代码

#include<stdio.h>
#define ll __int64
const int N=3e5+5;
int n,a[N],b[N],c[N],d[N],e[N];
ll ans,sum1,tail1,sum2,tail2;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
    {
        while(1<=tail1&&b[tail1]<=a[i])
        {
            sum1-=(d[tail1]-d[tail1-1])*1LL*b[tail1];
            tail1--;
        }
        while(1<=tail2&&c[tail2]>=a[i])
        {
            sum2-=(e[tail2]-e[tail2-1])*1LL*c[tail2];
            tail2--;
        }
        b[++tail1]=a[i];d[tail1]=i;
        c[++tail2]=a[i];e[tail2]=i;
        sum1+=(d[tail1]-d[tail1-1])*1LL*a[i];
        sum2+=(e[tail2]-e[tail2-1])*1LL*a[i];
        ans+=sum1-sum2;
    }
    printf("%I64d
",ans);
    return 0;
}