hdu 5443 The Water Problem 线段树

The Water Problem

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5443

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

 

HINT

 

题意

查询区间最大值，不带修改

题解：

随便怎么写，反正数据范围只有1000

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 10505
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

struct node
{
    int l,r;
    int ma;
};
node a[maxn*4];
int num[maxn];
void build(int x,int l,int r)
{
    a[x].l=l,a[x].r=r;
    if(l==r)
    {
        a[x].ma=num[l];
        return;
    }
    int mid=(l+r)>>1;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    a[x].ma=max(a[x<<1].ma,a[x<<1|1].ma);
}
int query(int x,int l,int r)
{
    int L=a[x].l,R=a[x].r;
    if(l<=L&&R<=r)
        return a[x].ma;
    int mid=(a[x].l+a[x].r)>>1;
    if(r<=mid)
        return query(x<<1,l,r);
    if(l>mid)
        return query(x<<1|1,l,r);
    return max(query(x<<1,l,mid),query(x<<1|1,mid+1,r));
}
int main()
{
    int t=read();
    while(t--)
    {
        memset(a,0,sizeof(a));
        int n=read();
        for(int i=1;i<=n;i++)
            num[i]=read();
        build(1,1,n);
        int q=read();
        for(int i=0;i<q;i++)
        {
            int l=read(),r=read();
            printf("%d\n",query(1,l,r));
        }
    }
}