VJ Balanced Lineup(ST表)

原题
一道很裸的RMQ，线段树专题里刷的，直接打ST表就好。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll f1[50005][20],f2[50005][20],a[50005];//二维数组别开太大，不然MLE了
int n,q,l,r;
void ST_prework()
{
    for(int i=1; i<=n; i++)
    {
        f1[i][0]=f2[i][0]=a[i];
    }
    for(int j=1; (1<<j)<=n; j++)
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            f1[i][j]=max(f1[i][j-1],f1[i+(1<<(j-1))][j-1]);
            f2[i][j]=min(f2[i][j-1],f2[i+(1<<(j-1))][j-1]);
        }
}
ll ST_querymax(int l,int r)
{
    int k=log(r-l+1)/log(2);
    return max(f1[l][k],f1[r-(1<<k)+1][k]);
}
ll ST_querymin(int l,int r)
{
    int k=log(r-l+1)/log(2);
    return min(f2[l][k],f2[r-(1<<k)+1][k]);
}
int main()
{
    cin>>n>>q;
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&a[i]);
    }
    ST_prework();
    while(q--)
    {
        ll x,y;
        scanf("%d%d",&l,&r);
        x=ST_querymax(l,r);
        y=ST_querymin(l,r);
        printf("%lld\n",x-y);
    }
    return 0;
}

标签：Lineup,20,50005,int,VJ,ST,Balanced,include,ll
来源： https://www.cnblogs.com/Pecoz/p/12469844.html