线段树之Poj3264

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 27508 Accepted: 12918Case Time Limit: 2000MS

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 andQ.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤B ≤N), representing the range of cows from A toB 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 31734251 54 62 2

Sample Output

630

#include <iostream>#include <stdio.h>#include <math.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MAXN 500005int ma[MAXN];int mi[MAXN];//建立线段树void build(int l,int r,int rt)//左孩子，右孩子，数组下标的索引{    if(l==r)    {        scanf("%d",&ma[rt]);        mi[rt]=ma[rt];        return;    }    int m=(l+r)>>1;//取中间数，右移一位相当于除以2    build(lson);    build(rson);    ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);    mi[rt]=min(mi[rt<<1],mi[rt<<1|1]);}//用线段数查找，复杂度logNint queryma(int L,int R,int l,int r,int rt){    if(L<=l&&R>=r)return ma[rt];//查询边界大于l,r,则返回l，r中最大的    int m=(l+r)>>1;    int ret=0;    if(L<=m)ret=max(ret,queryma(L,R,lson));//选出l，m中在L，R范围中最大的    if(R>m)ret=max(ret,queryma(L,R,rson));//选出r，m中在L，R范围中最大的    return ret;}int querymi(int L,int R,int l,int r,int rt){    if(L<=l&&R>=r)return mi[rt];    int m=(l+r)>>1;    int ret=100000000;    if(L<=m)ret=min(ret,querymi(L,R,lson));//选出l，m中在L，R范围中最大的    if(R>m)ret=min(ret,querymi(L,R,rson));    return ret;}int main(){    int n,q;    cin>>n>>q;    build(1,n,1);    while(q--)    {        int s,e;        scanf("%d%d",&s,&e);        printf("%d\n",queryma(s,e,1,n,1)-querymi(s,e,1,n,1));    }    return 0;}