POJ3264(线段树入门题)

Balanced Lineup%I64d & %I64u

Time Limit:%I64d & %I64u

Description

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

Sample Output

6
3
0

#include"cstdio"
#include"cstring"
#include"algorithm"
#define lson (rt<<1),l,mid
#define rson (rt<<1)|1,mid+1,r
#define gmid (a[rt].l+a[rt].r)>>1;
using namespace std;
const int MAXN=50005;
struct node{
    int l,r;
    int maxn,minn;
}a[MAXN*4];

void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    if(l==r)
    {
        scanf("%d",&a[rt].maxn);
        a[rt].minn=a[rt].maxn;
        return ;
    }
    
    int mid=gmid;
    build(lson);
    build(rson);
    a[rt].maxn=max(a[rt<<1].maxn,a[(rt<<1)|1].maxn);
    a[rt].minn=min(a[rt<<1].minn,a[(rt<<1)|1].minn);
}

int query(int rt,int l,int r,int op)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        if(op==1)    return a[rt].maxn;
        else return a[rt].minn;
    }
    
    int mid=gmid;
    
    if(r<=mid)    return query(rt<<1,l,r,op);
    else if(mid<l) return query((rt<<1)|1,l,r,op);
    else{
        if(op==1)    return max(query(rt<<1,l,mid,op),query((rt<<1)|1,mid+1,r,op));
        else return min(query(rt<<1,l,mid,op),query((rt<<1)|1,mid+1,r,op));
    }
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,1,n);
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int diff=query(1,l,r,1)-query(1,l,r,2);//1表示求最大值,2表示求最小值 
            printf("%d
",diff);
        }
    }
    
}

POJ3264(线段树入门题)

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