Balanced Lineup POJ - 3264（基础线段树）

最新推荐文章于 2019-05-06 21:10:08 发布

原创最新推荐文章于 2019-05-06 21:10:08 发布 · 369 阅读

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本文介绍了一个基于线段树实现的程序，该程序用于解决农场主约翰挑选连续范围内身高相近的奶牛参加飞盘游戏的问题。通过构建线段树，程序能够高效地查询任意指定范围内奶牛身高的最大值和最小值，进而计算出身高差。

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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

Sample Output

6
3
0

刚开始学习线段树，好像时间可以再优化（我不会）

这个题算是最基础的线段树了，没有更新，只有查询最大值和最小值。主要理解了一下线段树的作用过程。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn= 50005;
const int INF = 1e9;
int c[maxn*4];
struct Node
{
    int minn,maxx;
}segTree[maxn*4];
void build(int node , int begin ,int end)
{
    if(begin==end) segTree[node].maxx=segTree[node].minn=c[end];
    else
    {
        int mid=(begin+end) >> 1;
        build(node*2,begin,mid);
        build(node*2+1,mid+1,end);

        segTree[node].maxx = max(segTree[node*2].maxx,segTree[node*2+1].maxx);
        segTree[node].minn = min(segTree[node*2].minn,segTree[node*2+1].minn);
    }
}

int query(int node , int begin , int end, int left ,int right , int flag)
{
    int mid = (begin+end) >>1;
    if(flag)
    {
        if(left <=begin&&end<=right ) return segTree[node].maxx;
        if(left > end || right <begin ) return 0;

        int p1=query(node*2,begin,mid,left,right,flag);
        int p2=query(node*2+1,mid+1,end,left,right,flag);

        return p1>p2?p1:p2;
    }
    else
    {
        if(left <=begin&&end<=right ) return segTree[node].minn;
        if(left > end || right <begin ) return INF;

        int p1=query(node*2,begin,mid,left,right,flag);
        int p2=query(node*2+1,mid+1,end,left,right,flag);

        return p1<p2?p1:p2;
    }
}
int main()
{
    int n,q;
    cin >> n >> q;
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    build(1,1,n);
    for(int i=1;i<=q;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        cout << query(1,1,n,x,y,1)-query(1,1,n,x,y,0) <<endl;
    }
    return 0;
}