玲珑杯 Round 19 B Buildings （RMQ + 二分）

最新推荐文章于 2023-03-13 01:25:02 发布

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最新推荐文章于 2023-03-13 01:25:02 发布

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原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7258721.html

本文介绍了一种使用RMQ（Range Maximum Query）与二分搜索算法解决区间和谐性问题的方法。给定一系列建筑物的高度，需要计算满足特定条件的区间数量。通过预处理最大值和最小值的范围查询，有效地解决了该问题。

DESCRIPTION

There are

An inteval

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers

OUTPUT

Print a line of one number which means the answer.

SAMPLE INPUT

3 1 1 2 3

SAMPLE OUTPUT

HINT

Harmonious intervals are:

//Round 19 B;
#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll; 
int maxn[200005][25],minn[200005][25];
int a[200005];
int n,k,x;
void get_rmq()
{
    for(int i=1;i<=n;i++)
    {
        maxn[i][0]=a[i];
        minn[i][0]=a[i];
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)
        {
            maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
            minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
        }
    }
}
bool rmq(int l,int r)
{
    if(l>r) return 0;
    int f=int(log((double)(r-l+1))/log(2.0));
    return max(maxn[l][f],maxn[r-(1<<f)+1][f])-min(minn[l][f],minn[r-(1<<f)+1][f])<=k;
}
int main() 
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    get_rmq();
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        int l=1,r=i,mid,pos=i;
        while(l<=r)
        {
            mid=(l+r)>>1;
            if(rmq(mid,i)) r=mid-1,pos=mid;
            else l=mid+1;
        }
        ans+=i-pos+1;
    }
    printf("%lld\n",ans);
    return 0;
}