[POJ3264]Balanced Lineup（线段树，区间最值差）

题目链接：http://poj.org/problem?id=3264

写了一个更优美一点的

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>

using namespace std;

#define fr first
#define sc second
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a))

typedef struct Node {
    int lo, hi;
    Node() {}
    Node(int l, int h) : lo(l), hi(h) {}
}Node;

#define lrt rt << 1
#define rrt rt << 1 | 1
const int maxn = 500010;
Node h[maxn<<1];
int n, q;

void pushUP(int rt) {
    h[rt].lo = min(h[lrt].lo, h[rrt].lo);
    h[rt].hi = max(h[lrt].hi, h[rrt].hi);
}

void build(int l, int r, int rt) {
    if(l == r) {
        Rint(h[rt].lo);
        h[rt].hi = h[rt].lo;
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, lrt);
    build(m+1, r, rrt);
    pushUP(rt);
}

Node query(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) return h[rt];
    int m = (l + r) >> 1;
    Node ret(0x7f7f7f, 0), tmp;
    if(m >= L) {
        tmp = query(L, R, l, m, lrt);
        ret.lo = min(ret.lo, tmp.lo);
        ret.hi = max(ret.hi, tmp.hi);
    }
    if(m < R) {
        tmp = query(L, R, m+1, r, rrt);
        ret.lo = min(ret.lo, tmp.lo);
        ret.hi = max(ret.hi, tmp.hi);
    }
    return ret;
}

int main() {
    // FRead();
    int a, b;
    while(~Rint(n) && ~Rint(q)) {
        build(1, n, 1);
        while(q--) {
            Rint(a); Rint(b);
            Node ret = query(a, b, 1, n, 1);
            printf("%d
", ret.hi - ret.lo);
        }
    }
    return 0;
}