BZOJ1636&&1699：[USACO2007JAN]Balanced Lineup

浅谈\(RMQ\)：https://www.cnblogs.com/AKMer/p/10128219.html

题目传送门：https://lydsy.com/JudgeOnline/problem.php?id=1636

题目传送门：https://lydsy.com/JudgeOnline/problem.php?id=1699

裸的\(RMQ\)

时间复杂度：\(O(nlogn+m)\)

空间复杂度：\(O(nlogn)\)

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=5e4+5;

int n,m;
int a[maxn],Log[maxn];
int f[17][maxn],g[17][maxn];

int read() {
    int x=0,f=1;char ch=getchar();
    for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
    for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
    return x*f;
}

void make_st() {
    Log[0]=-1;
    for(int i=1;i<=n;i++)
        Log[i]=Log[i>>1]+1;
    for(int i=1;i<17;i++)
        for(int j=1;j+(1<<i)-1<=n;j++) {
            f[i][j]=max(f[i-1][j],f[i-1][j+(1<<(i-1))]);
            g[i][j]=min(g[i-1][j],g[i-1][j+(1<<(i-1))]);
        }
}

int query(int l,int r) {
    int x=Log[r-l+1];
    int mx=max(f[x][l],f[x][r-(1<<x)+1]);
    int mn=min(g[x][l],g[x][r-(1<<x)+1]);
    return mx-mn;
}

int main() {
    n=read(),m=read();
    for(int i=1;i<=n;i++)
        f[0][i]=g[0][i]=a[i]=read();
    make_st();
    for(int i=1;i<=m;i++) {
        int l=read(),r=read();
        printf("%d\n",query(l,r));
    }
    return 0;
}