poj3264 Balanced Lineup-优快云博客

本文链接：https://blog.youkuaiyun.com/Kirito_Acmer/article/details/45367715

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Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 37683		Accepted: 17656
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

Sample Output

6
3
0

这道题可以用线段树做，定义一个结构体，定义两个变量high,low,其中high记录这条线段中最大的点，low为最小的，每次更新到b[i].l==l&&b[i].r==r判断这段最大值以及最小值和原来所记录的相比怎么样。

#include<stdio.h>
#include<string.h>
#define inf 88888888
int max(int a,int b){
	return a>b?a:b;
}
int min(int a,int b){
	return a<b?a:b;
}


int low,high;
int a[50006];
struct node
{
	int l,r,high,low;
}b[4*50006];


void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;
	if(l==r){
		b[i].low=b[i].high=a[l];return;
	}
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
	b[i].high=max(b[i*2].high,b[i*2+1].high);
	b[i].low=min(b[i*2].low,b[i*2+1].low);
}


void question(int l,int r,int i)
{
	int mid;
	if(b[i].l==l && b[i].r==r){
		if(b[i].high>high)high=b[i].high;
		if(b[i].low<low)low=b[i].low;
		return;
	}
	mid=(b[i].l+b[i].r)/2;
	if(r<=mid)question(l,r,i*2);
	else if(l>mid)question(l,r,i*2+1);
	else {
		question(l,mid,i*2);question(mid+1,r,i*2+1);
	}
}


int main()
{
	int n,m,i,j,c,d;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		build(1,n,1);
		for(i=1;i<=m;i++){
			scanf("%d%d",&c,&d);
			low=inf;high=0;
			question(c,d,1);
			printf("%d\n",high-low);
		}
	}
	return 0;
}