Balanced Lineup POJ 3264（线段树）

http://poj.org/problem?id=3264

题意：现有N个数字，问你在区间[a, b]之间最大值与最小值的差值为多少？

分析：线段树模板，不过需要有两个查询，一个查询在该区间的最大值，一个查询在该区间的最小值，最后两者结果相减即可。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <math.h>

using namespace std;

#define met(a, b) memset(a, b, sizeof(a))
#define maxn 51000
#define INF 0x3f3f3f3f
const int MOD = 1e9+7;

typedef long long LL;
#define lson rt<<1
#define rson rt<<1|1
int A[maxn], n;

struct node
{
    int l, r, maxs, mins;
}tree[maxn<<2];

void build(int l, int r, int rt)
{
    tree[rt].l = l;
    tree[rt].r = r;

    if(l == r)
    {
        tree[rt].maxs = A[l];
        tree[rt].mins = A[l];
        return ;
    }

    int mid = (l+r)/2;
    build(l, mid, lson);
    build(mid+1, r, rson);

    tree[rt].maxs = max(tree[lson].maxs, tree[rson].maxs);
    tree[rt].mins = min(tree[lson].mins, tree[rson].mins);
}

int query1(int l, int r, int rt)
{
    if(tree[rt].l==l && tree[rt].r==r)
        return tree[rt].maxs;

    int mid = (tree[rt].l+tree[rt].r)/2;

    if(r<=mid) return query1(l, r, lson);
    else if(l>mid) return query1(l, r, rson);
    else
    {
        int lmaxs = query1(l, mid, lson);
        int rmaxs = query1(mid+1, r, rson);
        int maxs = max(lmaxs, rmaxs);

        return maxs;
    }
}

int query2(int l, int r, int rt)
{
    if(tree[rt].l==l && tree[rt].r==r)
        return tree[rt].mins;

    int mid = (tree[rt].l+tree[rt].r)/2;

    if(r<=mid) return query2(l, r, lson);
    else if(l>mid) return query2(l, r, rson);
    else
    {
        int lmins = query2(l, mid, lson);
        int rmins = query2(mid+1, r, rson);
        int mins = min(lmins, rmins);

        return mins;
    }
}

int main()
{
    int q, a, b;

    while(scanf("%d %d", &n, &q)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%d", &A[i]);

            build(1, n, 1);

        while(q --)
        {
            scanf("%d %d", &a, &b);

            int p = query1(a, b, 1);///查询最大值
            int q = query2(a, b, 1);///查询最小值

            printf("%d\n", p-q);
        }

    }
    return 0;
}