“玲珑杯”ACM比赛 Round #19 B.Buildings【二分+RMQ】

1149 - Buildings

Time Limit：2s Memory Limit：128MByte

Submissions：617Solved：159

DESCRIPTION

There are $n$ buildings lined up, and the height of the $i$-th house is $h_i$.

An inteval $[l,r]$$(l\le r)$ is harmonious if and only if $max(h_l,\dots ,h_r)-min(h_l,\dots ,h_r)\le k$.

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers$n(1\le n\le 2\times {10}^5),k(0\le k\le {10}^9)$.The second line contains $n$ integers $h_i(1\le h_i\le {10}^9)$.

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: $[1,1],[2,2],[3,3],[1,2],[2,3]$.

SOLUTION

“玲珑杯”ACM比赛 Round #19

题目大意：

给你一个长度为N的序列，求有多少个区间，使得区间内最大值-最小值<=k；

思路：

我们知道，如果我们确定了左端点L的话，随着右端点R的增加，会使得最大值越来越大，最小值越来越小，也就是差会越来越大，所以我们这里包含一个单调性。

我们可以O（n）枚举左端点，然后二分右端点然后RMQ预处理能够查询区间的最大值和最小值。

过程维护一个最终位子Pos，那么Pos-L+1就是当前左端点对答案的贡献。

Ac代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>using namespace std;int maxsum[250000][30];int minsum[250000][30];int a[250000];int n,k;void rmq_init(){    for(int j = 1; (1<<j) <= n; ++j)        for(int i = 1; i + (1<<j) - 1 <= n; ++i)        {            maxsum[i][j] = max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);            minsum[i][j] = min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);        }}int Get_cha(int l, int r){    int k = log2(r-l+1);    int Max = max(maxsum[l][k], maxsum[r-(1<<k)+1][k]);    int Min = min(minsum[l][k], minsum[r-(1<<k)+1][k]);    return Max - Min;}int main(){    while(~scanf("%d%d",&n,&k))    {        for(int i = 1; i <= n;++i)        {            scanf("%d",&a[i]);            maxsum[i][0] = a[i];             minsum[i][0] = a[i];        }        rmq_init();        long long  ans = 0;        int l , r;        for(int i = 1; i <= n; ++i)        {            int l=i;            int r=n;            int pos=-1;            while(r-l>=0)            {                int mid=(l+r)/2;                if(Get_cha(i,mid)<=k)                {                    pos=mid;                    l=mid+1;                }                else r=mid-1;            }            ans += pos - i+1;        }        printf("%lld\n",ans);    }}