B. Approximating a Constant Range

题目链接：http://codeforces.com/problemset/problem/602/B

题意：

给你一个相邻数差不超过 1 的序列，求最长子串的长度，满足子串中的最大值减最小值也不超过 1。

思路：

区间最大值，区间最小值分别用ST表去维护就可以了，然后之后去二分答案就好了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <math.h>
#include <set>
#include <vector>
#include <stack>
 
using namespace std;
const int maxn= 1e5 + 10;
 
int n;
int arr[maxn];
int dp[maxn][100];
int dp1[maxn][100];
 
int Min(int l,int r){
    int k = log(r-l+1)/log(2);
    return  min(dp[l][k],dp[r-(1<<k)+1][k]);
}
 
int Max(int l,int r){
    int k = log(r-l+1)/log(2);
    return  max(dp1[l][k],dp1[r-(1<<k)+1][k]);
}
 
bool check(int x){
    for (int i=1;i+x-1<=n;i++){
        if (Max(i,i+x-1)-Min(i,i+x-1)<=1)
            return true;
    }
    return false;
}
 
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        scanf("%d",&arr[i]);
    }
    for (int i=1;i<=n;i++){
        dp[i][0] = arr[i];
        dp1[i][0] = arr[i];
    }
    for (int j=1;(1<<j)<=n;j++){
        for (int i=1;i+(1<<j)-1<=n;i++){
            dp[i][j] = min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
            dp1[i][j] = max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
        }
    }
    int l=1,r=n+1,mid;
    int len = 1;
    while (l<=r){
        mid = (l+r)/2;
        if (check(mid)){
            len = max(mid,len);
            l = mid + 1;
        }
        else{
            r = mid-1;
        }
    }
    printf("%d
",len);
    return 0;
}