Poj 3264 balanced lineup (the maximum and minimum values of the number of line segments)

Link: http://poj.org/problem? Id = 3264

 

Time limit:5000 Ms		Memory limit:65536 K
Total submissions:32772		Accepted:15421
Case time limit:2000 ms

Description

For the daily milking, Farmer John'sNCows (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 shocould not differ too much in height.

Farmer John has made a listQ(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, NAnd 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 AAnd B(1 ≤ A≤ B≤ N), Representing the range of cows from ATo BIntrusive.

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


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The idea is also very clear, that is, get a new root [] array to find the minimum value, and set Minn to 0.

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#include <iostream>using namespace std;#define Maxx 50010int str[Maxx],root[Maxx<<2],root1[Maxx<<2];int n;int minn,maxx;void make_tree(int l,int r,int rt){    if(l == r)    {        root[rt]=str[l];        root1[rt]=str[l];        return ;    }    int mid=(l+r)/2;    make_tree(l,mid,rt*2);    make_tree(mid+1,r,rt*2+1);    root[rt]=max(root[rt*2],root[rt*2+1]);    root1[rt]=min(root1[rt*2],root1[rt*2+1]);}void update(int l,int r,int rt,int a,int b){    if(l == r && l == a)    {        root[rt]=b;        root1[rt]=b;        return ;    }    int mid=(l+r)/2;    if(a<=mid)        update(l,mid,rt*2,a,b);    else        update(mid+1,r,rt*2+1,a,b);    root[rt]=max(root[rt*2],root[rt*2+1]);    root1[rt]=min(root1[rt*2],root1[rt*2+1]);}void query(int l,int r,int rt,int left,int right){    if(l == left&&r == right)    {        maxx=max(maxx,root[rt]);        minn=min(minn,root1[rt]);        return ;    }    int mid=(l+r)/2;    if(left>mid)    {        query(mid+1,r,rt*2+1,left,right);    }    else if(right<=mid)    {        query(l,mid,rt*2,left,right);    }    else    {        query(l,mid,rt*2,left,mid);        query(mid+1,r,rt*2+1,mid+1,right);    }}int main(){    int q;    int i,j;    int a,b;    while(scanf("%d%d",&n,&q)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&str[i]);        }        make_tree(1,n,1);        for(i=1;i<=q;i++)        {            minn=100000000;maxx=0;            scanf("%d%d",&a,&b);            query(1,n,1,a,b);            printf("%d\n",maxx-minn);        }    }    return 0;}

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