Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 45243		Accepted: 21240
Case Time Limit: 2000MS

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.

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.

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.

6 3
1
7
3
4
2
5
1 5
4 6
2 2

6
3
0

USACO 2007 January Silver

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分析 既可以用线段树也可以用ST来做

也是我的ST第一道题，不错

这道题我两种方法都写过的，比较简单，当做练熟练度

线段树的解法已经写过的，可以很明显的看出，ST的代码比线段树简洁得多

就直接上ST的代码

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #define maxx 50005
 5 using namespace std;
 6 int maxsum[maxx][18],minsum[maxx][18],n,Q;
 7 int _max(int a,int b)
 8 {
 9     return a>b?a:b;
10 }
11 int _min(int a,int b)
12 {
13     return a<b?a:b;
14 }
15 int main()
16 {
17
18     while(~scanf("%d %d\n",&n,&Q))
19     {
20         for(int i=1;i<=n;i++)
21         {
22             scanf("%d\n",&maxsum[i][0]);
23             minsum[i][0]=maxsum[i][0];
24         }
25         for(int j=1;j<=18;j++)// 预处理
26         for(int i=1;i<=n;i++)    //j 循环在 i 循环外
27         {
28             if(i+(1<<j)-1<=n)        // 注意左右区间
29             {
30                 maxsum[i][j]=_max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);
31                 minsum[i][j]=_min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);
32             }
33         }
34         while(Q--)
35         {int a,b;
36             scanf("%d %d\n",&a,&b);
37             int k=(int)((log((double)(b-a+1)))/log(2.0));    //注意要包含端点并转换成double
38             int maxl=_max(maxsum[a][k],maxsum[b-(1<<k)+1][k]);//不然有些测试要出问题 比如poj
39             int minl=_min(minsum[a][k],minsum[b-(1<<k)+1][k]);
40             printf("%d\n",maxl-minl);
41         }
42     }
43     return 0;
44 }