玲珑杯”ACM比赛 Round #19 B 维护单调栈

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;

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <cstdio>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <cctype>
 #include <map>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <string>
 #include <complex>
 #define ll long long
 #define mod 1000000007
 using namespace std;
 ll n,k;
 ll a[];
 ll l[],r[];
 map<ll,map<ll,ll>> mp;
 int main()
 {
         scanf("%lld %lld",&n,&k);
         for(int i=; i<=n; i++)
             scanf("%lld",&a[i]);
         a[]=-1ll;
         a[n+]=-1ll;
         l[]=1ll;
         for(int i=; i<=n; i++) //关键********
         {
             ll temp=i-;
             while(a[temp]>=a[i])//维护一个递增的序列
                 temp=l[temp]-;
             l[i]=temp+;
         }
         r[n]=n;
         for (int i=n-; i>=; i--)
         {
             ll temp=i+;
             while(a[temp]>=a[i])
                 temp=r[temp]+;
             r[i]=temp-;
         }
         ll ans=;
         for(int i=; i<=n; i++)
         {
             ll x=,y=;
             if(mp[l[i]][r[i]]!=){//去重 更改l[i]
               ll now=l[i];
               l[i]=mp[l[i]][r[i]];
               mp[now][r[i]]=i+;
              }
              else
             mp[l[i]][r[i]]=i+;
             for(int j=i-; j>=l[i]; j--)
             {
                 if((a[j]-a[i])<=k)
                     x++;
                 else
                     break;
             }
             for(int j=i+; j<=r[i]; j++)
             {
                 if((a[j]-a[i])<=k)
                     y++;
                 else
                     break;
             }
             ans=ans+x*y+x+y+1ll;
         }
         printf("%lld\n",ans);
     return ;
 }