玲珑OJ 1149 - Buildings 【尺取法】

题目链接

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显然通过尺取法可以找到每次以j为结尾,包含j的区间,往左延伸,最左能到i (i<=j)
于是以j为结尾的区间有(j-i+1)个
累加即可

#include<stdio.h>
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MEM(a,x) memset(a,x,sizeof(a))
#define lowbit(x) ((x)&-(x))

using namespace std;

const int N = 2e5+5;

int a[N];

map<int,int>mp;

void eraseAi(int i){
    auto it=mp.find(a[i]);
    --it->second;
    if(it->second==0){
        mp.erase(it);
    }
}

ll slove(int n,int k){
    mp.clear();
    int i=0;
    mp[a[i]]=1;
    ll ans=1;
    for(int j=1;j<n;++j){
        mp[a[j]]+=1;
        while((--mp.end())->first-mp.begin()->first>k){
            eraseAi(i);
            ++i;
        }
        ans+=(j-i+1);
    }
    return ans;
}

int main()
{
    //freopen("/home/lu/code/r.txt","r",stdin);
    //freopen("/home/lu/code/w1.txt","w",stdout);
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;++i){
            scanf("%d",&a[i]);
        }
        printf("%lld\n",slove(n,k));
    }
    return 0;
}