Neat Tree（线段树）所有区间的最大值-最小值的和

链接： https://www.nowcoder.com/acm/contest/106/I
来源：牛客网

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

It’s universally acknowledged that there’re innumerable trees in the campus of HUST.

There is a row of trees along the East-9 Road which consists of N trees. Now that we know the height of each tree, the gardeners of HUST want to know the super-neatness of this row of trees. The neatness of a sequence of trees is defined as the difference between the maximum height and minimum height in this sequence. The super-neatness of this sequence of trees is defined as the sum of neatness of all continous subsequences of the trees.

Multiple cases please process until the end of input. There are at most 100 test cases.

输入描述:

For each case:
The first line contains an integer N represents the number of

trees.

The next line n positive integers followed, in which h _i $(1\leq h_{i} \leq 10^{6})$ represent the

height of the tree i.

输出描述:

For each test case, output a integer in a single line which represents the super-neatness of the row
of trees.

示例1

输入

输出

6
10

说明

As for the first case, the super-neatness of the row of trees [1, 3, 1] is 6, because there are 6 different subsequence of this row of trees: [1] (from index 0 to index 0), neatness is 0; [1, 3] (from index 0 to index 1), neatness is 2; [1, 3, 1] (from index 0 to index 2), neatness is 2; [3] (from index 1 to index 1), neatness is 0; [3, 1] (from index 1 to index 2), neatness is 2; [1] (from index 2 to index 2), neatness is 0.

等有机会再去重新学一遍线段树吧，要深入的学习，不能浅尝即止，这是别人ac的代码，个人觉得这篇代码写的很规范,值得我去学习。

#include <cstdio>
#include <iostream>
#define N 1100000
using namespace std;
struct node
{
    int num, numid;
}t[4 * N];
typedef long long LL;
LL ans;
int n, h[N], p, q;
 
 
void buildmax(int x, int l, int r)
{
  int mid = (l + r) / 2;
  if (l == r)
  {
    t[x].num = h[l];
    t[x].numid = l;
    return;
  }
  buildmax(x * 2, l, mid);
  buildmax(x * 2 + 1, mid + 1, r);
  int ls = x * 2, rs = x * 2 + 1;
    t[x].num = max(t[ls].num, t[rs].num);
    if (t[x].num == t[ls].num)
        t[x].numid = t[ls].numid;
    else
        t[x].numid = t[rs].numid;
}
 
void buildmin(int x, int l, int r)
{
  int mid = (l + r) / 2;
  if (l == r)
  {
    t[x].num = h[l];
    t[x].numid = l;
    return;
  }
  buildmin(x * 2, l, mid);
  buildmin(x * 2 + 1, mid + 1, r);
  int ls = x * 2, rs = x * 2 + 1;
    t[x].num = min(t[ls].num, t[rs].num);
    if (t[x].num == t[ls].num)
        t[x].numid = t[ls].numid;
    else
        t[x].numid = t[rs].numid;
}
 
void querymax(int x, int l, int r, int ls, int rs)
{
    if (r < ls || l > rs) return;
    if (l <= ls && rs <= r)
    {
        if (t[x].num > p)
        {
            p = t[x].num;
            q = t[x].numid;
        }
        return;
    }
    int mid = (ls + rs) / 2;
    querymax(x * 2, l, r, ls, mid);
    querymax(x * 2 + 1, l, r, mid + 1, rs);
}
 
void querymin(int x, int l, int r, int ls, int rs)
{
    if (r < ls || l > rs) return;
    if (l <= ls && rs <= r)
    {
        if (t[x].num < p)
        {
            p = t[x].num;
            q = t[x].numid;
        }
        return;
    }
    int mid = (ls + rs) / 2;
    querymin(x * 2, l, r, ls, mid);
    querymin(x * 2 + 1, l, r, mid + 1, rs);
}
 
 
void askmax(int l, int r)
{
    if (l > r) return;
    p = 0;
    querymax(1, l, r, 1, n);
    int t = q;
//  cout << l << ' ' << r << ' ' << t << endl;
    ans += LL(t - l + 1) * (r - t + 1) * h[t];
    askmax(l, t - 1);
    askmax(t + 1, r);
}
 
void askmin(int l, int r)
{
    if (l > r) return;
    p = 100000000;
    querymin(1, l, r, 1, n);
    int t = q;
    ans -= LL(t - l + 1) * (r - t + 1) * h[t];
    askmin(l, t - 1);
    askmin(t + 1, r);
}
 
 
 
int main()
{
    //freopen("1.in", "r", stdin);
    //freopen("1.out", "w", stdout);
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i ++) scanf("%d", &h[i]);
        buildmax(1, 1, n);
 
      ans = 0;
      askmax(1, n);
      //cout << ans << endl;
      buildmin(1, 1, n);
      askmin(1, n);
      cout << ans << endl;
    }
    return 0;
}