【BZOJ 1636】 [Usaco2007 Jan]Balanced Lineup

1636: [Usaco2007 Jan]Balanced Lineup

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 516  Solved: 374
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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

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

Sample Input

* 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 Output

6
3
0

HINT

Source

Silver

RMQ水题。

预处理出max,min的RMQ数组，对于每个询问找max-min即可。

#include <iostream>#include <cmath>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;int maxx,minn,n,q,a[50005],mi[50005][20],ma[50005][20];void RMQ(){for (int i=1;i<=n;i++)mi[i][0]=ma[i][0]=a[i];for (int j=1;(1<<j)<n;j++)for (int i=1;i+(1<<j)-1<=n;i++)mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]),ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);}int Getlog(int x){int now=1;if (x==1) return 0;for (int i=1;i<=100;i++){now*=2;if (x==now||x<now*2) return i;}}void Ask(int l,int r){int k=Getlog(r-l+1);maxx=max(ma[l][k],ma[r-(1<<k)+1][k]);minn=min(mi[l][k],mi[r-(1<<k)+1][k]);}int main(){    scanf("%d%d",&n,&q);for (int i=1;i<=n;i++)scanf("%d",&a[i]);RMQ();while (q--){int x,y;scanf("%d%d",&x,&y);Ask(x,y);printf("%d\n",maxx-minn);}return 0;}

感悟：

好久没写过RMQ，但还清晰的记得怎么写；

但是前两天刚写过树套树，今天写就出了一堆错误。。

充分说明一个问题：对算法一定要理解透彻才能减少错误！！！