A - Balanced Lineup POJ - 3264

思路

• 题意

1. RMQ问题

• 分析

2. 用ST表 或 线段树

3. 算法讲解传送门

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <queue>
#include <map>
/* #include <unordered_map> */
#include <bitset>
#include <vector>
void fre() { system("clear"), freopen("A.txt", "r", stdin); freopen("Ans.txt","w",stdout); }
void Fre() { system("clear"), freopen("A.txt", "r", stdin);}
#define ios ios::sync_with_stdio(false)
#define Pi acos(-1)
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define db double
#define Pir pair<int, int>
#define PIR pair<Pir, Pir>
#define m_p make_pair
#define INF 0x3f3f3f3f
#define esp 1e-7
#define mod (ll)(1e9 + 7)
#define for_(i, s, e) for(int i = (ll)(s); i <= (ll)(e); i ++)
#define rep_(i, e, s) for(int i = (ll)(e); i >= (ll)(s); i --)
#define sc scanf
#define pr printf
#define sd(a) scanf("%d", &a)
#define ss(a) scanf("%s", a)
using namespace std;

const int mxn = 5e4 + 10;
int ar[mxn];
int st1[mxn][30], st2[mxn][30];
int Log[mxn]; 
void ST(int n)
{
    Log[1] = 0;
    for_(i, 2, n) Log[i] = Log[i / 2] + 1;
    for_(i, 1, n) st1[i][0] = st2[i][0] = ar[i];
    for(int j = 1; (1 << j) <= n; j ++)
        for(int i = 1; i + (1 << j) - 1 <= n; i ++)
        {
            st1[i][j] = max(st1[i][j - 1], st1[i + (1 << (j - 1))][j - 1]);
            st2[i][j] = min(st2[i][j - 1], st2[i + (1 << (j - 1))][j - 1]);
        }
}

int main()
{
    /* fre(); */
    int n, q;
    sc("%d %d", &n, &q);
    for_(i, 1, n)
        sd(ar[i]);
    ST(n);
    int l, r;
    while(q --)
    {
        sc("%d %d", &l, &r);
        int k = Log[r - l + 1];
        int mx = max(st1[l][k], st1[r - (1 << k) + 1][k]);
        int mn = min(st2[l][k], st2[r - (1 << k) + 1][k]);
        printf("%d\n", mx - mn);
    }

    return 0;
}