POJ 3264 Balanced Lineup（ST模板）

链接：http://poj.org/problem?id=3264

题意：给n个数，求一段区间L,R的最大值 - 最小值，Q次询问

思路：ST表模板，预处理区间最值，O（1）复杂度询问

AC代码：

#include<iostream>
#include<vector>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 50005;
const int logn = 21;
int Log[maxn];
int MIN[maxn][logn];
int MAX[maxn][logn];
int cow[maxn];
int n,q;
void pre(){
    Log[1] = 0;
    Log[2] = 1;
    for(int i = 3;i<=n;i++){
        Log[i] = Log[i/2] + 1;
    }
}
void st(){
    for(int i = 1;i<=n;i++){
        MIN[i][0] = MAX[i][0] = cow[i]; 
    }
    for(int j = 1;(1<<j)<=n;j++){
        for(int i = 1;i+(1<<j)-1<=n;i++){
            MAX[i][j] = max(MAX[i][j-1],MAX[i+(1<<(j-1))][j-1]);
            MIN[i][j] = min(MIN[i][j-1],MIN[i+(1<<(j-1))][j-1]); 
        }
    }
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i = 1;i<=n;i++){
        scanf("%d",&cow[i]);
    }
    pre();
    st();
    while(q--){
        int A,B;
        scanf("%d%d",&A,&B);
        int s = Log[B-A+1];
        int range_max = max(MAX[A][s],MAX[B-(1<<s)+1][s]);
        int range_min = min(MIN[A][s],MIN[B-(1<<s)+1][s]);
        cout<<range_max-range_min<<endl;
    }
    return 0;
}