POJ 3264 Balanced Lineup

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 57016 Accepted: 26718
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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题意：
给一个序列，静态查询区间最大值和最小值的差。

思路：
这题是学st表(Sparse Table)的入门题。虽然区间查询可以用树状数组或者线段树，但是st表真好写真好写真好写啊……
接下来是st表的介绍……

对于一个序列，定义st[i][j]为从第i个元素开始2j个元素中的最大(小)值。
以最小值为例：st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1])

int stMin[maxn][50];void st_prepare(){    for(int i=n-1; i>=0; i--)    {        stMax[i][0]=stMin[i][0]=a[i];        for(int j=1; i+(1<<j)-1 < n; j++)        {            stMin[i][j]=min(stMin[i][j-1],stMin[i+(1<<(j-1))][j-1]);        }    }}

访问区间 (l,r) 的最小值，令k=[log2(len)]。
最小值为min(stMin[l][k],stMin[r-(1<< k)+1][k])。

代码：

#include<cstdio>#include<iostream>#define maxn 100000using namespace std;int n,q;int a[maxn];int stMin[maxn][50];int stMax[maxn][50];void st_prepare(){    for(int i=n-1; i>=0; i--)    {        stMax[i][0]=stMin[i][0]=a[i];        for(int j=1; i+(1<<j)-1 < n; j++)        {            stMax[i][j]=max(stMax[i][j-1],stMax[i+(1<<(j-1))][j-1]);            stMin[i][j]=min(stMin[i][j-1],stMin[i+(1<<(j-1))][j-1]);        }    }}/*int queryMax(int l,int r){    int k;    int len=r-l+1;    for(k=0;k<50;k++)    {        if((1<<k)<=len&&(1<<(k+1))>=len) break;    }    return max(stMax[l][k],stMax[r-(1<<k)+1][k]);}int queryMin(int l,int r){    int k;    int len=r-l+1;    for(k=0;k<50;k++)    {        if((1<<k)<=len&&(1<<(k+1))>=len) break;    }    return min(stMin[l][k],stMin[r-(1<<k)+1][k]);}*/int query(int l,int r){    int k;    int len=r-l+1;    for(k=0;k<50;k++)    {        if((1<<k)<=len&&(1<<(k+1))>=len) break;    }    return max(stMax[l][k],stMax[r-(1<<k)+1][k]) - min(stMin[l][k],stMin[r-(1<<k)+1][k]);}int main(){    while(~scanf("%d %d",&n,&q))    {        int i,j;        for(i=0;i<n;i++) scanf("%d",&a[i]);        st_prepare();        while(q--)        {            int l,r;            scanf("%d %d",&l,&r);            printf("%d\n",query(l-1,r-1));        }    }}