POJ3264 Balanced Lineup RMQ/线段树

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 23931		Accepted: 11122
Case 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 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

/* 功能Function Description:     POJ-3264
   开发环境Environment:          DEV C++ 4.9.9.1
   技术特点Technique:
   版本Version:
   作者Author:                   可笑痴狂
   日期Date:                      20120815
   备注Notes:                    ------RMQ算法(O(1)的时间复杂度查询区间最值)
   题意：
        查询区间最大值与最小值之差
*/

/*
//代码一：-----RMQ
#include<stdio.h>
#include<math.h>
#define MAX_N 50000+10
#define MAX_L 16            //2^16>50000

//定义存储的都是下标
int maxN[MAX_N][MAX_L]; //maxN[i][j]存储的是从编号i开始的连续2^j个数中最大值的下标
int minN[MAX_N][MAX_L]; //minN[i][j]存储的是从编号i开始的连续2^j个数中最小值的下标
int h[MAX_N];

int max(int i,int j)
{
    return h[i]>h[j]?i:j;
}

int min(int i,int j)
{
    return h[i]<h[j]?i:j;
}

void preST(int n)    //ST的预处理
{
    int i,j;
    for(i=0;i<n;++i)        //初始时候 h[minN[i][0]]=h[maxN[i][0]]=h[i]
        minN[i][0]=maxN[i][0]=i;
    for(j=1;(1<<j)<=n;++j)  //dp
    {
        for(i=0;i+(1<<j)-1<n;++i)
        {
            maxN[i][j]=max(maxN[i][j-1],maxN[i+(1<<(j-1))][j-1]);
            minN[i][j]=min(minN[i][j-1],minN[i+(1<<(j-1))][j-1]);
        }
    }
}

int ST(int a,int b,int flag)
{
    int pow;
    --a;
    --b;
    for(pow=1;(1<<pow)<b-a+1;++pow);
    --pow;
    //pow=(int)log(b-a+1)/log(2);  //直接计算pow但是poj不支持  (pow是使2^pow<b-a+1成立的最大值，只有这样才能把a到b拆成两段而完全覆盖)
    if(flag)    //返回最大值
        return h[max(maxN[a][pow],maxN[b-(1<<pow)+1][pow])];   //因为前后两段有重叠，所以注意是从后边向前计算边界
    else        //返回最小值
        return h[min(minN[a][pow],minN[b-(1<<pow)+1][pow])];
}

int main()
{
    int n,m,a,b,i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;++i)
            scanf("%d",&h[i]);
        preST(n);
        for(i=0;i<m;++i)
        {
            scanf("%d%d",&a,&b);
            printf("%d\n",ST(a,b,1)-ST(a,b,0));
        }
    }
    return 0;
}
*/

//代码二：-----------线段树
#include<stdio.h>
#include<stdlib.h>
#define MAX 50000
int ansmin,ansmax;

struct node
{
    int max;
    int min;
    int lc;
    int rc;
}tree[MAX*3];

void build(int l,int r,int T)
{
    int mid=(l+r)>>1;
    tree[T].lc=l;
    tree[T].rc=r;
    tree[T].min=0x3fffffff;
    tree[T].max=-1;
    if(l==r)
        return;
    else
    {
        build(l,mid,T<<1);
        build(mid+1,r,(T<<1)|1);
    }
}

void insert(int num,int k,int T)
{
    int mid=(tree[T].lc+tree[T].rc)>>1;
    if(num<tree[T].lc||num>tree[T].rc)
        return;
    if(tree[T].max<k)
        tree[T].max=k;
    if(tree[T].min>k)
        tree[T].min=k;
    if(tree[T].lc==tree[T].rc)
        return;
    if(num<=mid)
        insert(num,k,T<<1);
    else
        insert(num,k,(T<<1)|1);
}

void qurry(int s,int t,int T)
{
    int mid=(tree[T].lc+tree[T].rc)>>1;
    if(tree[T].min>ansmin&&tree[T].max<ansmax)  //剪枝
        return;
    if(tree[T].lc==s&&tree[T].rc==t)
    {
        if(ansmin>tree[T].min)  //这里要用if语句判断一下，否则在分区间寻找的时候不能综合考虑两个区间的最值
            ansmin=tree[T].min;
        if(ansmax<tree[T].max)
            ansmax=tree[T].max;
        return;
    }
    if(t<=mid)
        qurry(s,t,T<<1);
    else if(s>mid)
        qurry(s,t,(T<<1)|1);
    else
    {
        qurry(s,mid,T<<1);
        qurry(mid+1,t,(T<<1)|1);
    }
}

int main()
{
    int n,m,i,a,b,t;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,n,1);
        for(i=1;i<=n;++i)
        {
            scanf("%d",&t);
            insert(i,t,1);
        }
        for(i=0;i<m;++i)
        {
            ansmin=0x3fffffff;
            ansmax=-1;
            scanf("%d%d",&a,&b);
            qurry(a,b,1);
            printf("%d\n",ansmax-ansmin);
        }
    }
    return 0;
}