首页 > 代码库 > POJ3264_Balanced Lineup(线段树/单点更新)

POJ3264_Balanced Lineup(线段树/单点更新)

2024-07-16 09:08:08   225人阅读

解题报告

题意：

求区间内最大值和最小值的差值。

思路：

裸线段树，我的线段树第一发。区间最值。

#include <iostream>
#include <cstring>
#include <cstdio>
#define  inf 99999999
#define LL long long
using namespace std;
LL minn[201000],maxx[201000];
void update(LL root,LL l,LL r,LL p,LL v)
{
    if(l==r)minn[root]=v,maxx[root]=v;
    if(l<r)
    {
        LL mid=(l+r)/2;
        if(p<=mid)update(root*2,l,mid,p,v);
        else update(root*2+1,mid+1,r,p,v);
        minn[root]=min(minn[root*2],minn[root*2+1]);
        maxx[root]=max(maxx[root*2],maxx[root*2+1]);
    }
}
LL q_minn(LL root,LL l,LL r,LL ql,LL qr)
{
    LL mid=(l+r)/2,ans=inf;
    if(ql<=l&&r<=qr)return minn[root];
    if(ql<=mid)ans=min(ans,q_minn(root*2,l,mid,ql,qr));
    if(qr>mid)ans=min(ans,q_minn(root*2+1,mid+1,r,ql,qr));
    return ans;
}
LL q_maxx(int root ,int l,int r,int ql,int qr)
{
    LL mid=(l+r)/2,ans=-inf;
    if(ql<=l&&r<=qr)return maxx[root];
    if(ql<=mid)ans=max(ans,q_maxx(root*2,l,mid,ql,qr));
    if(mid<qr)ans=max(ans,q_maxx(root*2+1,mid+1,r,ql,qr));
    return ans;
}
int main()
{
    LL a,n,i,j,q,ql,qr;
    scanf("%lld%lld",&n,&q);
    for(i=1; i<=n; i++)
    {
        scanf("%lld",&a);
        update(1,1,n,i,a);
    }
    while(q--)
    {
        scanf("%lld%lld",&ql,&qr);
        printf("%lld\n",q_maxx(1,1,n,ql,qr)-q_minn(1,1,n,ql,qr));
    }
    return 0;
}

Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 34181 Accepted: 16065
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


联系
我们