ST表 || RMQ问题 || BZOJ 1699: [Usaco2007 Jan]Balanced Lineup排队 || Luogu P2880 [USACO07JAN]平衡的阵容Balanced Lineup

题面：P2880 [USACO07JAN]平衡的阵容Balanced Lineup

题解：

ST表板子

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int maxn=50050,max_log=18,maxlog=17,inf=1<<30;
int Fmx[maxn][max_log],N,Q,Fmn[maxn][max_log],LOG[maxn],A,B;
int main(){
    scanf("%d%d",&N,&Q);
    for(int i=0;i<=N;i++)
        for(int j=0;j<=maxlog;j++)
            Fmn[i][j]=inf;
    for(int i=1;i<=N;i++){
        scanf("%d",&Fmx[i][0]);
        Fmn[i][0]=Fmx[i][0];
    }
    LOG[0]=-1;
    for(int i=1;i<=N;i++)LOG[i]=LOG[i>>1]+1;
    for(int j=1;j<=maxlog;j++)//
        for(int i=1;i+(1<<j)-1<=N;i++){
            Fmx[i][j]=max(Fmx[i][j-1],Fmx[i+(1<<(j-1))][j-1]);
            Fmn[i][j]=min(Fmn[i][j-1],Fmn[i+(1<<(j-1))][j-1]);
        }
    while(Q--){
        scanf("%d%d",&A,&B);
        int x=LOG[B-A+1],ans1,ans2;
        ans1=max(Fmx[A][x],Fmx[B-(1<<x)+1][x]);
        ans2=min(Fmn[A][x],Fmn[B-(1<<x)+1][x]);
        printf("%d
",ans1-ans2);
    }
    return 0;
}

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By:AlenaNuna