Poj.3264.Balanced Lineup（线段树）

Poj.3264.Balanced Lineup

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

题意：给定一序列，求指定区间上最大值与最小值之差

考点：线段树

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

const int N=50007;
struct st                          //Segment Tree：每个点代表一条线段
{
    int left,right;                //非元线段的左右孩子
    int mina;                      //线段上的最大最小值
    int maxa;
};
st q[N*4];                         //开4*N的数组以免越界
int n,m,ans1,ans2;

void build(int num,int L,int R)    //构造线段树:节点个数,左边界,右边界
{
    q[num].left=L;
    q[num].right=R;
    if(L==R)
    {
        scanf("%d",&q[num].maxa);
        q[num].mina=q[num].maxa;
        return;
    }
    build(2*num,L,(L+R)/2);                                      //左孩子区间
    build(2*num+1,(L+R)/2+1,R);                                  //右孩子区间
    q[num].mina=min(q[2*num].mina,q[2*num+1].mina);
    q[num].maxa=max(q[2*num].maxa,q[2*num+1].maxa);
}

int get_max(int num,int s,int t)                                 //区间[s,t]上最大值
{
    if(q[num].left>=s&&q[num].right<=t) 
        return q[num].maxa;
    int mid=(q[num].left+q[num].right)/2;
    if(mid>=t) 
        return get_max(2*num,s,t);
    else if(mid<s) 
        return get_max(2*num+1,s,t);
    else 
        return max(get_max(2*num,s,mid),get_max(2*num+1,mid+1,t));
}

int get_min(int num,int s,int t)                                 //区间[s,t]上最小值
{
    if(q[num].left>=s&&q[num].right<=t)
        return q[num].mina;
    int mid=(q[num].left+q[num].right)/2;
    if(mid>=t)
        return get_min(2*num,s,t);
    else if(mid<s)
        return get_min(2*num+1,s,t);
    else
        return min(get_min(2*num,s,mid),get_min(2*num+1,mid+1,t));
}

int main()
{
    scanf("%d%d",&n,&m);                                          //n个节点,m次询问
    build(1,1,n);                                                 //构造线段树
    while(m--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",get_max(1,a,b)-get_min(1,a,b));
    }
    return 0;
}