“玲珑杯”ACM比赛 Round #19 B(RMQ大法好啊,比线段树快得多!!!!)

分析:水题。枚举一个左区间,然后二分一个满足的右区间。算贡献即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[200010];
int maxl[200010*2][30];
int minl[200010*2][30];
int n,kk;
void S_table()
{
    int l = int(log((double)n)/log(2.0));
    for (int j=1;j<=l;j++)
    {
        for (int i=1; i + (1 << (j-1) ) - 1 <=n;++i)
        {
            maxl[i][j] = max(maxl[i][j-1], maxl[i + (1 << (j-1) )][j-1]);
            minl[i][j] = min(minl[i][j-1], minl[i + (1 << (j-1) )][j-1]);
        }
    }
}
int rmq(int l, int r)
{
    int k = int(log((double)(r-l+1))/log(2.0));
    int a1 = max(maxl[l][k], maxl[r - (1<<k) + 1][k]);
    int a2 = min(minl[l][k], minl[r - (1<<k) + 1][k]);
    return a1-a2<=kk;
}
int main()
{
    scanf("%d%d",&n,&kk);
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&a[i]);
        maxl[i][0]=minl[i][0]=a[i];
    }
    S_table();
    ll ans=n;
    for(int i=1;i<n;i++)
    {
        int l=i,r=n;int pos=i;
        while(r>=l)
        {
            int mid=(l+r)>>1;
            if(rmq(i,mid))pos=mid,l=mid+1;
            else r=mid-1;
        }
        ans+=pos-i;
    }
    printf("%lld\n",ans);
    return 0;
}
    原文作者:B树
    原文地址: https://blog.csdn.net/nucshiyilang/article/details/76339474
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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