HDU5443——The Water Problem(2015年长春区域赛)_2015 区域赛题目最短路-优快云博客

本文链接：https://blog.youkuaiyun.com/LDUtyk/article/details/52506322

解决水源查询问题，通过遍历指定区间的水源大小，找到并返回最大水源。适用于资源有限的情况。

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The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1715 Accepted Submission(s): 1363

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

Sample Output

Source

2015 ACM/ICPC Asia Regional Changchun Online

题意：
寻找水源，给定n个水源，每个水源可用谁为ai，给定一个区间，求这个区间内的最大水源。

解：暴力寻找，水源水量放在数组中，下标作为位置，在给定位置区间中暴力寻找最大水源。

#include<stdio.h>
#include<string.h>

int a[1005];

int main()
{
	int t;
	int n,q;
	int l,r;
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			int max=0;
			memset(a,0,sizeof(a));
			scanf("%d",&n);
			for(int i=1;i<=n;i++){
				scanf("%d",&a[i]);
			}
			scanf("%d",&q);
			for(int j=0;j<q;j++){
				max=0;
				scanf("%d%d",&l,&r);
				for(int z=l;z<=r;z++){
					if(max<a[z])
						max=a[z];
				}
				printf("%d\n",max);
			}
		}
	}
 }