poj 3264 线段树求区间最大最小值

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

Source

USACO 2007 January Silver

线段树处理该问题存储区间最大值与最小值

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#define INF 0xfffffff
using namespace std;
struct tree
{
    int L;
    int R;
    int minv;
    int maxv;
    int mid()
    {
        return (L+R)/2;
    }
}tree[800010];
int maxv=-INF;
int minv=INF;
void buildtree(int root,int L,int R)
{
    tree[root].L=L;
    tree[root].R=R;
    tree[root].minv=INF;
    tree[root].maxv=-INF;
    if(L!=R)
    {
        buildtree(2*root+1,L,(L+R)/2);
        buildtree(2*root+2,(L+R)/2+1,R);
    }
}
void inse(int root,int i,int v)
{
    if(tree[root].L==tree[root].R)
    {
        tree[root].minv=v;
        tree[root].maxv=v;
        return ;
    }
    tree[root].minv=min(tree[root].minv,v);
    tree[root].maxv=max(tree[root].maxv,v);
    if(i<=tree[root].mid())
        inse(2*root+1,i,v);
    else
        inse(2*root+2,i,v);
}
void query(int root ,int s,int e)
{
    if(tree[root].minv>minv&&tree[root].maxv<maxv)
        return ;
    if(tree[root].L==s&&tree[root].R==e)
    {
        minv=min(minv,tree[root].minv);
        maxv=max(maxv,tree[root].maxv);
        return ;
    }
    if(e<=tree[root].mid())
        query(2*root+1,s,e);
    else if(s>tree[root].mid())
        query(2*root+2,s,e);
    else
    {
        query(2*root+1,s,tree[root].mid());
        query(2*root+2,tree[root].mid()+1,e);
    }
}
int main()
{
    int n,q;
    int re;
    int ss,ee;
        scanf("%d%d",&n,&q);
        buildtree(0,1,n);
        for(int i=1; i<=n; i++)
        {
            //cout<<"**********"<<endl;
            scanf("%d",&re);
            inse(0,i,re);
        }
        for(int j=1; j<=q; j++)
        {
            scanf("%d%d",&ss,&ee);
            minv=INF;
            maxv=-INF;
            query(0,ss,ee);
            printf("%d
",maxv-minv);
        }

return 0;
}

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poj 3264 线段树 求区间最大最小值

poj 3264 线段树求区间最大最小值