洛谷 2880 平衡的阵容（RMQ）

最新推荐文章于 2021-09-11 15:55:32 发布

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本文介绍了一道USACO算法题“平衡的阵容”的解决思路，该问题需要从连续的牛群中选择高度差异最小的一组进行游戏。文章详细解释了如何使用倍增RMQ算法快速求解区间最大值和最小值，从而高效地计算每组牛的最大和最小高度差。

P2880 [USACO07JAN]平衡的阵容Balanced Lineup

题目描述

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.

一个农夫有N头牛，每头牛的高度不同，我们需要找出最高的牛和最低的牛的高度差。

输入输出格式

输入格式：
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.

输出格式：
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.

输入输出样例

输入样例#1：
6 3
1
7
3
4
2
5
1 5
4 6
2 2
输出样例#1：
6
3
0

很明显的一道倍增RMQ，注意一下端点的问题就可以了

代码如下

program mys;

var i,j,k,m,mi,ma,n,t,x1,y1,q:longint;
a:array[0..100000]of longint;
two:array[0..20]of longint;
f,g:array[0..150000,0..16]of longint;

function min(a,b:longint):longint;
begin 
if a<b then exit(a)
else exit(b);
end;

function max(a,b:longint):longint;
begin 
if a>b then exit(a)
else exit(b);
end;

begin 
readln(n,q);
fillchar(g,sizeof(g),$7f);
for i:=1 to n do 
begin 
readln(a[i]);
f[i,0]:=a[i];
g[i,0]:=a[i];
end;

two[0]:=1;
for i:=1 to 20 do 
two[i]:=two[i-1]*2;

for j:=1 to 16 do 
for i:=1 to n do 
begin 
f[i,j]:=max(f[i,j-1],f[i+two[j-1],j-1]);
g[i,j]:=min(g[i,j-1],g[i+two[j-1],j-1]);
end;

for i:=1 to q do 
begin 
readln(x1,y1);
if x1=y1 then 
begin 
writeln(0);
continue;
end;
t:=trunc(ln(y1-x1+1)/ln(2));
mi:=maxlongint;
mi:=min(g[x1,t],g[y1-two[t]+1,t]);
ma:=max(f[x1,t],f[y1-two[t]+1,t]);
writeln(abs(ma-mi));
end;
end.