Balanced Lineup

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

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

HINT

题意

给你n个数，然后查询区间的最大值减去最小值是多少

题解：

RMQ，我用ST来做，也可线段树。

注意：　j+1<<(i-1) 和　j+(1<<(i-1)) 是不一样的，位运算优先级还不如加法。

代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 #define N 200050
 7 int n,m,mi[N][21],mx[N][21];
 8 template<typename T>void read(T &x)
 9 {
10   int k=0; char c=getchar();
11   x=0;
12   while(!isdigit(c)&&c!=EOF)c^=c=='-',c=getchar();
13   if(c==EOF)exit(0);
14   while(isdigit(c))x=x*10+c-'0',c=getchar();
15   x=k?-x:x;
16 }
17 int query_max(int x,int y)
18 {
19   int k=(int)(log(y-x+1)/log(2));
20   return max(mx[x][k],mx[y-(1<<k)+1][k]);
21 }
22 int query_min(int x,int y)
23 {
24   int k=(int)(log(y-x+1)/log(2));
25   return min(mi[x][k],mi[y-(1<<k)+1][k]);
26 }
27 int main()
28 {
29   #ifndef ONLINE_JUDGE
30   freopen("aa.in","r",stdin);
31   #endif
32   read(n);read(m);
33   for(int i=1;i<=n;i++)read(mx[i][0]),mi[i][0]=mx[i][0];
34   for(int i=1;i<=20;i++)
35     for(int j=1;j<=n;j++)
36       if (j+(1<<(i-1))>n)break;
37       else
38     {
39       mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);
40       mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1]);
41     }
42   for(int i=1;i<=m;i++)
43     {
44       int x,y;
45       read(x);read(y);
46       int ans=query_max(x,y)-query_min(x,y);
47       printf("%d\n",ans);
48     }
49 }

View Code

poj 3264 Balanced Lineup RMQ问题