time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A function is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all . We’ll deal with a more… discrete version of this term.
For an array , we define it’s Lipschitz constant as follows:
if n < 2,
if n ≥ 2, over all 1 ≤ i < j ≤ n
In other words, is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.
You are given an array of size n and q queries of the form [l, r]. For each query, consider the subarray ; determine the sum of Lipschitz constants of all subarrays of .
Input
The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array and the number of queries respectively.
The second line contains n space-separated integers ().
The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).
Output
Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of .
Examples
input
10 4
1 5 2 9 1 3 4 2 1 7
2 4
3 8
7 10
1 9
output
17
82
23
210
input
7 6
5 7 7 4 6 6 2
1 2
2 3
2 6
1 7
4 7
3 5
output
2
0
22
59
16
8
Note
In the first query of the first sample, the Lipschitz constants of subarrays of with length at least 2 are:
The answer to the query is their sum.
【题目链接】:http://codeforces.com/contest/602/problem/D
【题解】
题意:
给你n个数字;
然后给你q个区间
每个区间l,r
让你求出[l,r]这个区间里面的数字的所有子列的L(h)的值的和;
(L(h)是任意两个点的斜率的绝对值的最大值);
做法:
可以肯定让你求L..R这个子列的区间的L(h)值
L(h)值必然是在abs(a[i]-a[i-1])中取到(i∈【L..R-1】)
可以看下图;
假设斜率的最大值不在相邻的两点之间得到;
设中间还有一个k;
则如果ak大于ai，则ai,ak连起来肯定更优,肯定比ai,aj连起来的斜率大
如果ak小于ai的话，则是ak,aj连起来更优;
总之只有在i,j相邻的时候,斜率才可能取到最大;
这样的话,我们就先把L..R这个区间内的相邻的数的差的绝对值处理出来
变为bi数组
然后枚举第bi中的第i个数字为斜率的最大值
看看这第i个数字能够“管理”“统治”的区间范围;
然后这段区间里面只要包括这第i个数字,则最大值肯定是就是第i个数字;
设其左边有l[i]个数字，右边有r[i]个数字
则答案递增l[i]*r[i]*b[i];
(左边选取l[i]个数字中的一个作为左端点,右边选取r[i]中的一个数字作为右端点,两个点组合成一个包围第i个点的区间);
这种问题可以用单调队列来搞;(单调栈?)
但是要注意重复的问题,所以左边在判断的时候加等号
右边判断的时候不加等号;
这样就不会重复计数了;
(可以拿样例中的第二组询问来理解这个重复区间的问题);
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int MAXN = 1e5+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n,q,top;
LL l[MAXN],r[MAXN],b[MAXN],a[MAXN];
int s[MAXN];
int main()
{
//freopen("F:\rush.txt","r",stdin);
rei(n);rei(q);
rep1(i,1,n)
rel(a[i]);
rep1(i,1,q)
{
LL L,R,len,ans = 0;
rel(L);rel(R);
len = R-L;
rep1(j,L,R-1)
b[j-L+1] = abs(a[j+1]-a[j]);
//j∈L..R-1
top=0;
for (LL j = 1;j <= len;j++)
{
while (top && b[j]>=b[s[top]]) top--;//加等号
if (top==0)
l[j] = 0;
else
l[j] = s[top];
//s[top]>b[j]
s[++top] = j;
}
top = 0;
for (LL j = len;j >=1 ;j--)
{
while (top && b[j]>b[s[top]]) top--;//不加等号
if (top==0)
r[j] = len+1;
else
r[j] = s[top];
//s[top]>b[j]
s[++top] = j;
}
for (LL j = 1;j <= len;j++)
ans+=(j-l[j])*(r[j]-j)*b[j];
cout << ans << endl;
}
return 0;
}