POJ 3264 Balanced Lineup 線段樹入門(點的查詢)

畫船聽雨發表於2014-02-20

這道題的意思是：給你n個點，每個數出現的位置對應著改點數的大小。然後給出m個區間輸出每個區間上的最大值與最小值的差。

典型的線段樹啊，自己寫的那個程式碼很搓，然後嘯爺教了我，他怎麼寫的，就學習了一下。以後儘量寫的好一點啊。

主要是線段樹的建立與查詢。

Balanced Lineup
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 31222   Accepted: 14709
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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 50100;
using namespace std;

struct node
{
    int a, b;
    int Max, Min;
} f[4*maxn];

int cnum[maxn];


void Bulid(int l, int r, int site)
{
    if(l == r)
    {
        f[site].a = l;
        f[site].b = r;
        f[site].Max = cnum[l];
        f[site].Min = cnum[l];
        return ;
    }
    int mid = (l+r)/2;
    Bulid(l, mid, site*2);
    Bulid(mid+1, r, site*2+1);
    f[site].Max = max(f[site*2].Max, f[site*2+1].Max);
    f[site].Min = min(f[site*2].Min, f[site*2+1].Min);
    f[site].a = l;
    f[site].b = r;
}

node Search(int l, int r, int site)
{
    if(l == f[site].a && r == f[site].b)
        return f[site];
    int mid = (f[site].a+f[site].b)/2;
    if(r <= mid)
    {
        return Search(l, r, 2*site);
    }
    else if(l > mid)
    {
        return Search(l, r, site*2+1);
    }
    else
    {
        node p1 = Search(l, mid, site*2);
        node p2 = Search(mid+1, r, site*2+1);
        node p;
        p.Max = max(p1.Max, p2.Max);
        p.Min = min(p1.Min, p2.Min);
        return p;
    }
}


void Updata(int m, int d, int site)
{
    if(f[site].a == f[site].b)
    {
        f[site].Max = f[site].Min = d;
        return;
    }
    int mid = (f[site].a+f[site].b)/2;
    if(m <= mid)
    {
        Updata(m, d, 2*site);
    }
    else
    {
        Updata(m, d, 2*site+1);
    }
    f[site].Max = max(f[site*2].Max, f[site*2+1].Max);
    f[site].Min = min(f[site*2].Min, f[site*2+1].Min);
}

int main()
{
    int n, m;
    while(scanf("%d %d",&n, &m) != EOF)
    {
        for(int i = 1; i <= n; i++)
            scanf("%d",&cnum[i]);
        Bulid(1, n, 1);
        int a, b;
        while(m--)
        {
            scanf("%d %d",&a, &b);
            node p;
            p = Search(a, b, 1);
            printf("%d\n",p.Max-p.Min);
        }
    }
    return 0;
}


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