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 $\mathrm{nn\; buildings\; lined\; up,\; and\; the\; height\; of\; the\; ii-th\; house\; is\; hihi.}$

An inteval $[l,r][l,r](l\le r)(l\le r)\; is\; harmonious\; if\; and\; only\; if\; max(hl,\dots ,hr)-min(hl,\dots ,hr)\le kmax(hl,\dots ,hr)-min(hl,\dots ,hr)\le k.$

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers $\mathrm{n(1\le n\le 2\times 105),k(0\le k\le 109)n(1\le n\le 2\times 105),k(0\le k\le 109).\; The\; second\; line\; contains\; nn\; integers\; hi(1\le hi\le 109)hi(1\le hi\le 109).}$

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[200005];
ll l[200005],r[200005];
map<ll,map<ll,ll>> mp;
int main()
{
        scanf("%lld %lld",&n,&k);
        for(int i=1; i<=n; i++)
            scanf("%lld",&a[i]);
        a[0]=-1ll;
        a[n+1]=-1ll;
        l[1]=1ll;
        for(int i=2; i<=n; i++) //关键********
        {
            ll temp=i-1;
            while(a[temp]>=a[i])//维护一个递增的序列
                temp=l[temp]-1;
            l[i]=temp+1;
        }
        r[n]=n;
        for (int i=n-1; i>=1; i--)
        {
            ll temp=i+1;
            while(a[temp]>=a[i])
                temp=r[temp]+1;
            r[i]=temp-1;
        }
        ll ans=0;
        for(int i=1; i<=n; i++)
        {
            ll x=0,y=0;
            if(mp[l[i]][r[i]]!=0){//去重 更改l[i]
              ll now=l[i];
              l[i]=mp[l[i]][r[i]];
              mp[now][r[i]]=i+1;
             }
             else
            mp[l[i]][r[i]]=i+1;
            for(int j=i-1; j>=l[i]; j--)
            {
                if((a[j]-a[i])<=k)
                    x++;
                else
                    break;
            }
            for(int j=i+1; j<=r[i]; j++)
            {
                if((a[j]-a[i])<=k)
                    y++;
                else
                    break;
            }
            ans=ans+x*y+x+y+1ll;
        }
        printf("%lld
",ans);
    return 0;
}