1149 - Buildings
Time Limit：2s Memory Limit：128MByte
Submissions：588Solved：151
DESCRIPTION
There are nn buildings lined up, and the height of the ii-th house is hihi.
An inteval [l,r][l,r](l≤r)(l≤r) is harmonious if and only if max(hl,…,hr)−min(hl,…,hr)≤kmax(hl,…,hr)−min(hl,…,hr)≤k.
Now you need to calculate the number of harmonious intevals.
INPUT
The first line contains two integers n(1≤n≤2×105),k(0≤k≤109)n(1≤n≤2×105),k(0≤k≤109). The second line contains nn integers hi(1≤hi≤109)hi(1≤hi≤109).
OUTPUT
Print a line of one number which means the answer.
SAMPLE INPUT
3 1
1 2 3
SAMPLE OUTPUT
5
HINT
Harmonious intervals are: [1,1],[2,2],[3,3],[1,2],[2,3][1,1],[2,2],[3,3],[1,2],[2,3].
题意：给你一个长度为n的序列 问有多少区间 使得 区间最大值-区间最小值<=k
题解：单调栈处理出以a[i]为最小值的区间左界右界 组合出合法的区间 注意 （同一左界右界)或者称为同一块 下的最小值可能会有重复
从左向右遍历时 将当前值的左界改为(同一块中上一个相同值的位置+1) 具体看代码;
1 #pragma comment(linker, "/STACK:102400000,102400000")
2 #include <cstdio>
3 #include <iostream>
4 #include <cstdlib>
5 #include <cstring>
6 #include <algorithm>
7 #include <cmath>
8 #include <cctype>
9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <bitset>
13 #include <string>
14 #include <complex>
15 #define ll long long
16 #define mod 1000000007
17 using namespace std;
18 ll n,k;
19 ll a[200005];
20 ll l[200005],r[200005];
21 map<ll,map<ll,ll>> mp;
22 int main()
23 {
24 scanf("%lld %lld",&n,&k);
25 for(int i=1; i<=n; i++)
26 scanf("%lld",&a[i]);
27 a[0]=-1ll;
28 a[n+1]=-1ll;
29 l[1]=1ll;
30 for(int i=2; i<=n; i++) //关键********
31 {
32 ll temp=i-1;
33 while(a[temp]>=a[i])//维护一个递增的序列
34 temp=l[temp]-1;
35 l[i]=temp+1;
36 }
37 r[n]=n;
38 for (int i=n-1; i>=1; i--)
39 {
40 ll temp=i+1;
41 while(a[temp]>=a[i])
42 temp=r[temp]+1;
43 r[i]=temp-1;
44 }
45 ll ans=0;
46 for(int i=1; i<=n; i++)
47 {
48 ll x=0,y=0;
49 if(mp[l[i]][r[i]]!=0){//去重 更改l[i]
50 ll now=l[i];
51 l[i]=mp[l[i]][r[i]];
52 mp[now][r[i]]=i+1;
53 }
54 else
55 mp[l[i]][r[i]]=i+1;
56 for(int j=i-1; j>=l[i]; j--)
57 {
58 if((a[j]-a[i])<=k)
59 x++;
60 else
61 break;
62 }
63 for(int j=i+1; j<=r[i]; j++)
64 {
65 if((a[j]-a[i])<=k)
66 y++;
67 else
68 break;
69 }
70 ans=ans+x*y+x+y+1ll;
71 }
72 printf("%lld\n",ans);
73 return 0;
74 }