poj3264 balanced lineup【线段树】

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q. 
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i 
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

两棵树分别保存区间最大最小值即可

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 1e18
using namespace std;

int q, n;
const int maxn = 50005;
int cow[maxn], talltree[maxn << 2], shorttree[maxn<<2];

void tall_pushup(int rt)//更新
{
    talltree[rt] = max(talltree[rt << 1], talltree[rt << 1 | 1]);
}

void short_pushup(int rt)
{
    shorttree[rt] = min(shorttree[rt<<1], shorttree[rt<<1|1]);
}

void tall_build(int l, int r, int rt)
{
    if(l == r){
        talltree[rt] = cow[l];
        return;
    }
    int m = (l + r) >> 1;
    tall_build(l, m, rt << 1);
    tall_build(m + 1, r, rt << 1 | 1);
    tall_pushup(rt);
}

void short_build(int l, int r, int rt)
{
    if(l == r){
        shorttree[rt] = cow[l];
        return;
    }
    int m = (l + r) >> 1;
    short_build(l, m, rt << 1);
    short_build(m + 1, r, rt << 1 | 1);
    short_pushup(rt);
}

/*void pushdown(int rt, int ln, int rn)
{
    if(lazy[rt]){
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        tree[rt << 1] += lazy[rt] * ln;
        tree[rt << 1 | 1] += lazy[rt] * rn;
        lazy[rt] = 0;
    }
}

void update(int L, int C, int l, int r, int rt)
{
    if(l == r){
        tree[rt] += C;
        return;
    }
    int m = (l + r) >>1;
    if(L <= m) update(L, C, l, m, rt << 1);
    else update(L, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}

void update(int L, int R, int C, int l, int r, int rt)
{
    if(L <= l && r <= R){//本区间完全在操作区间内
        tree[rt] += C * (r - l + 1);
        lazy[rt] += C;
        return;
    }
    int m = (l + r) >> 1;
    pushdown(rt, m - l + 1, r - m);
    if(L <= m) update(L, R, C, l, m, rt << 1);
    if(R > m) update(L, R, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}*/

int tall_query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R){
        return talltree[rt];
    }
    int m = (l + r) >> 1;
    //pushdown(rt, m - l + 1, r - m);

    int ans = 0;
    if(L <= m) ans = max(ans, tall_query(L, R, l, m, rt << 1));
    if(R > m) ans = max(ans, tall_query(L, R, m + 1, r, rt << 1 | 1));
    return ans;
}

int short_query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R){
        return shorttree[rt];
    }
    int m = (l + r) >> 1;
    //pushdown(rt, m - l + 1, r - m);

    int ans = inf;
    if(L <= m) ans = min(ans, short_query(L, R, l, m, rt << 1));
    if(R > m) ans = min(ans, short_query(L, R, m + 1, r, rt << 1 | 1));
    return ans;
}


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

        while(q--){
            int a, b;
            scanf("%d%d", &a, &b);
            cout<<tall_query(a, b, 1, n, 1) - short_query(a, b, 1, n, 1)<<endl;
        }
    }
    return 0;
}