HDOJ 5443 The Water Problem（线段树区间最值）（长春网络赛）

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 302    Accepted Submission(s): 245

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


    
    
    
    很明显的区间最大值 ac代码： 
    
    
    
    
#include<stdio.h>
struct s
{
	int left;
	int right;
	int max;
}tree[1000*4];
int MAX(int a,int b)
{
	return a>b?a:b;
}
void build(int i,int l,int r)
{
	tree[i].left=l;
	tree[i].right=r;
	if(l==r)
	{
		int num;
		scanf("%d",&num);
		tree[i].max=num;
		return;
	}
	int mid=(l+r)/2;
	build(i*2,l,mid);
	build(i*2+1,mid+1,r);
	tree[i].max=MAX(tree[i*2].max,tree[i*2+1].max);
}
int query(int l,int r,int i)
{
	if(l<=tree[i].left&&r>=tree[i].right)
	{
		return tree[i].max;
	}
	int mid=(tree[i].left+tree[i].right)/2;
	if(r<=mid)
	return query(l,r,i*2);
	if(l>mid)
	return query(l,r,i*2+1);
	return MAX(query(l,mid,i*2),query(mid+1,r,i*2+1));
}
int main()
{
	int t;
	int a,b,i,n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		build(1,1,n);
		scanf("%d",&m);
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			printf("%d\n",query(a,b,1));
		}
	}
	return 0;
}