POJ - 3264 Balanced Lineup

题目链接

线段树，求区间最大值减去区间最小值的值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int N = 5e4 + 10;
const int inf = 1e9 + 7;
int n,q;

#define mid ((l + r) >> 1)
#define ls (nod << 1)
#define rs (nod << 1 | 1)
#define lson ls,l,mid
#define rson rs,mid + 1,r

int mx[N << 2],mi[N << 2];
void build(int nod,int l,int r){
    if(l == r) { scanf("%d",&mx[nod]); mi[nod] = mx[nod]; return ; }
	build(lson); build(rson);
	mx[nod] = max(mx[ls],mx[rs]); mi[nod] = min(mi[ls],mi[rs]);
}
int query(int nod,int l,int r,int ll,int rr,int pos){
    if(l > rr || r < ll) return pos ? -inf : inf;
    if(ll <= l && r <= rr) return pos ? mx[nod] : mi[nod];
    return pos ? max(query(lson,ll,rr,pos), query(rson,ll,rr,pos)) : min(query(lson,ll,rr,pos), query(rson,ll,rr,pos));
}

int main(){
    scanf("%d%d",&n,&q);
    build(1,1,n);
    while(q --){
	    int l,r; scanf("%d%d",&l,&r);
	    printf("%d\n",query(1,1,n,l,r,1) - query(1,1,n,l,r,0));
	}
	return 0;
}