hdu5443 The Water Problem（水）

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,...,anrepresenting 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

区间最大值   线段树

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const int oo = 0x3f3f3f3f;
const int N = 1005;
typedef long long LL;
struct da
{
    int left, right, val;
}as[N*4];
int n, ac[N];
void build(int left, int right, int i)
{
    as[i].left = left;
    as[i].right = right;
    int mid = (left + right)/2;
    if(left == right)
    {
        as[i].val = ac[left];
        return ;
    }
    build(left, mid, 2*i);
    build(mid+1, right, 2*i+1);
    as[i].val = max(as[i*2].val, as[2*i+1].val);
}
void query(int left, int right, int i, int &ans)
{
    if(as[i].left == left && as[i].right == right)
    {
        ans = max(as[i].val, ans);
        return ;
    }
    int mid = (as[i].left + as[i].right)/2;
    if(right <= mid)  query(left, right, 2*i, ans);
    else if(left > mid) query(left, right, 2*i+1, ans);
    else
    {
        query(left, mid, 2*i, ans);
        query(mid+1, right, 2*i+1, ans);
    }
}
int main()
{
    int T, Q;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%d", &ac[i]);
        build(1, n, 1);
        scanf("%d", &Q);
        while(Q--)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            int ans = 0;
            query(a, b, 1, ans);
            printf("%d\n", ans);
        }
    }
    return 0;
}

　　

 

转载于:https://www.cnblogs.com/PersistFaith/p/4808128.html