POJ 3264 Balanced Lineup（最大最小差值线段树水题）

最新推荐文章于 2018-08-21 14:50:09 发布

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最新推荐文章于 2018-08-21 14:50:09 发布

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分类专栏：线段树文章标签：线段树

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 37122		Accepted: 17383
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

Source

USACO 2007 January Silver

最大最小差值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 50005

struct stud{
int le,ri;
int mi,ma;
}f[N*4];

int a[N],mi,ma;

void build(int pos,int le,int ri)
{
	f[pos].le=le;
	f[pos].ri=ri;
	if(le==ri)
	{
		f[pos].mi=f[pos].ma=a[le];
		return ;
	}

    int mid=MID(le,ri);
    build(L(pos),le,mid);
    build(R(pos),mid+1,ri);

    f[pos].ma=max(f[L(pos)].ma,f[R(pos)].ma);
    f[pos].mi=min(f[L(pos)].mi,f[R(pos)].mi);
}

void query(int pos,int le,int ri)
{
	if(f[pos].le==le&&f[pos].ri==ri)
	{
		mi=min(mi,f[pos].mi);
		ma=max(ma,f[pos].ma);
		return ;
	}

   int mid=MID(f[pos].le,f[pos].ri);

   if(mid>=ri)
	query(L(pos),le,ri);
   else
	if(mid<le)
	 query(R(pos),le,ri);
   else
   {
   	 query(L(pos),le,mid);
   	 query(R(pos),mid+1,ri);
   }

}

int main()
{
	int i,j;
	int n,m;
	while(~sff(n,m))
	{
		fre(i,1,n+1)
		 sf(a[i]);

		 build(1,1,n);
		 int le,ri;
		 while(m--)
		 {
		 	sff(le,ri);
		 	mi=INF;
		 	ma=-1;
            query(1,le,ri);
		 	pf("%d\n",ma-mi);
		 }
	}
   return 0;
}