【HDU】 5443 线段树

最新推荐文章于 2020-08-23 13:49:15 发布

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本文介绍了一个基于线段树实现的区间最大值查询算法，适用于数据范围不大于1000的情况。通过构建线段树，可以高效地进行多次查询，找到指定区间内的最大水源大小。

Problem Description
In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.
 

Input
First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an, and each integer is in {1,...,106}. On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.
 

Output
For each query, output an integer representing the size of the biggest water source.
 

Sample Input
3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3
 

Sample Output
100
2
3
4
4
5
1
999999
999999
1

题意描述：

查询区间最大值，不带修改

题解：
数据范围1000

写个线段树吧暴力也可以的

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

const int INF=-0x3f3f3f3f;
const int MAXN=200000;

struct st{
    int left,right;
    int maxx;
};
st q[MAXN*4];
int n,m;


void built(int num,int L,int R)
{
    q[num].left=L;
    q[num].right=R;
    if(q[num].left==q[num].right)
    {
        scanf("%d",&q[num].maxx);
        return;
    }
    built(2*num,L,(L+R)/2);
    built(2*num+1,(L+R)/2+1,R);
    q[num].maxx=max(q[2*num].maxx,q[2*num+1].maxx);
}

int get_maxa(int s,int t,int num)
{
    if(s<=q[num].left&&t>=q[num].right)
        return q[num].maxx;
    if(s>q[num].right||t<q[num].left)
        return INF;
    int a,b;
    a=get_maxa(s,t,2*num);
    b=get_maxa(s,t,2*num+1);
    return max(a,b);
}

int main()
{
    int n,qq;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int s,t;
        scanf("%d",&n);
        built(1,1,n);
        scanf("%d",&qq);
        while(qq--)
        {
            scanf("%d%d",&s,&t);
            printf("%d\n",get_maxa(s,t,1));
        }
    }
    return 0;
}