Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 66277		Accepted: 30847
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

Source

USACO 2007 January Silver

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題意： 給你一組資料有q個詢問求l,r之間的最大值和最小值得差值

題解：rmq一個存小值一個存大值

---

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
#define clr(a,b)  memset(a,b,sizeof(a))
#define il        inline
typedef long long ll;
const int maxn = 50000 + 2;
const int minn = 130 + 1;

int n,q,dpmax[maxn][20],dpmin[maxn][20];
void RMQ(int n)
{
	for (int j = 0; j < 17; ++j)
	{
		for (int i = n; i > 0; --i)
		{
			if (i + (1 << j) <= n)
				{
				    dpmax[i][j + 1] = max(dpmax[i][j], dpmax[i + (1 << j)][j]);
				    dpmin[i][j + 1] = min(dpmin[i][j], dpmin[i + (1 << j)][j]);
				}
		}
	}
}
int querymax(int l, int r)
{
	if (l > r) { return -1; }
	int k=(int)(log(r-l+1.0)/log(2.0));
	return max(dpmax[l][k], dpmax[r - (1 << k) + 1][k]);
}
int querymin(int l, int r)
{
	if (l > r) { return -1; }
	int k=(int)(log(r-l+1.0)/log(2.0));
	return min(dpmin[l][k], dpmin[r - (1 << k) + 1][k]);
}
int main()
{
	while (~scanf("%d%d", &n,&q))
	{
		clr(dpmax, 0);
		clr(dpmin, 0);
		int tm;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &tm);
			dpmax[i][0]=dpmin[i][0]=tm;
		}
		RMQ(n);
		int l,r;
		while(q--)
        {
            scanf("%d%d",&l,&r);
            printf("%d\n",querymax(l,r)-querymin(l,r));
        }
	}
	return 0;
}