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 ≤ 180,000) 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.

每天,农夫 John 的 n(1≤n≤5×104)n(1\le n\le 5\times 10^4)n(1≤n≤5×104) 头牛总是按同一序列排队。

有一天, John 决定让一些牛们玩一场飞盘比赛。他准备找一群在对列中为置连续的牛来进行比赛。但是为了避免水平悬殊,牛的身高不应该相差太大。John 准备了 q(1≤q≤1.8×105)q(1\le q\le 1.8\times10^5)q(1≤q≤1.8×105) 个可能的牛的选择和所有牛的身高 hi(1≤hi≤106,1≤i≤n)h_i(1\le h_i\le 10^6,1\le i\le n)hi(1≤hi≤106,1≤i≤n)。他想知道每一组里面最高和最低的牛的身高差。

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.

第一行两个数 n,q。

接下来 n行，每行一个数 hi。

再接下来 q 行，每行两个整数 a 和 b，表示询问第 a 头牛到第 b头牛里的最高和最低的牛的身高差。

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.

输出共 q 行，对于每一组询问，输出每一组中最高和最低的牛的身高差。

输入输出样例

输入 #1

输出 #1

6
3
0

解法一：

树状数组：建树时维护最大最小值，而不是求前缀和。其次，最大最小值要通过查询分开找，因此有两个查询函数find_max(x,y)和find_min(x,y).

1. y−lowbit(y)>x ，这种情况下，我们可以把求[x,y]区间的最值拆分成两部分，先求[x,y−lowbit(y)][中最值与[y−lowbit(y)+1,y]中的最值，再求这两者的最值。

我们发现 [y−lowbit(y)+1,y][y−lowbit(y)+1,y] 就是tree[y]）。

于是，问题就转化为：求 [x,y−lowbit(y)]中最值与 tree[y]的最值。

2.y−lowbit(y)<x，在这种情况下，我们直接把 a[y]（第y个输入的数据）给剥离出来，于是原问就变成了：求 [x,y−1]中最值与 a[y] 的最值。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e7;
int n,q,h[maxn],trmax[maxn],trmin[maxn];
int lowbit(int x){//lowbit运算 return x&(-x);
}
void build(int x, int k){//插入建树 for(;x<=n;x+=lowbit(x)){trmax[x]=max(trmax[x],k);//不断维护 trmin[x]=min(trmin[x],k);}
}
int find_max(int x, int y){//查询最大值 if(y>x){if(y-lowbit(y)>x) return max(trmax[y],find_max(x,y-lowbit(y)));else return max(h[y],find_max(x,y-1));}return h[x];
}
int find_min(int x, int y){//查询最小值 if(y>x){if(y-lowbit(y)>x){return min(trmin[y],find_min(x,y-lowbit(y)));}else return min(h[y],find_min(x,y-1));}return h[x];
}
int main(){memset(trmin,0x3f3f3f3f,sizeof(trmin));//初始化trmin为INF memset(trmax,0,sizeof(trmax));//初始化trmax为0 cin>>n>>q;for(int i=1;i<=n;i++){cin>>h[i];build(i,h[i]);//建树 }while(q--){int x,y;cin>>x>>y;cout<<find_max(x,y)-find_min(x,y)<<endl;}return 0;
}

解法二：ST算法倍增复杂度：O（Nlog(N))

ST:设f[i][j]表示数列a中下标在子区间[i,i+2^j-1]里的数的最大值，递推边界是f[i][0]=a[i]
递推时，我们把子区间的长度成倍增长，有公式f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);

模板如下：

void prework(){for(int i=1;i<=n;i++) f[i][0]=a[i];int t=log(n)/log(2)+1;for(int j=1;j<t;j++){for(int i=1;i<=n-(1<<j)+1;i++){f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);}}
}
int ST_query(int l, int r){int k=log(r-l+1)/log(2);return max(f[l][k],f[r-(1<<k)+1][k]);
}

代码如下：


#include <bits/stdc++.h>
using namespace std;
int n,q;
int ma[180010][22],mi[180010][21];//记录最大值，最小值
int ST(int l, int r){int t=log(r-l+1)/log(2);int x,y;//板子 x=max(ma[l][t],ma[r-(1<<t)+1][t]);y=min(mi[l][t],mi[r-(1<<t)+1][t]);return x-y;//最大值减最小值
}
int main(){cin>>n>>q;for(int i=1;i<=n;i++){cin>>ma[i][0];mi[i][0]=ma[i][0];//初始化 f[i][0]=a[i] }int t=log(n)/log(2)+1;for(int j=1;j<t;j++){for(int i=1;i<=n-(1<<j)+1;i++){//板子 ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);}}while(q--){int x,y;cin>>x>>y;cout<<ST(x,y)<<endl;}return 0;
}

解法三：线段树（然鹅我不会，等我会了再来更吧，hhh）

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题目描述

输入格式

输出格式

输入输出样例

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