hdu 5443 裸st表

最新推荐文章于 2021-04-29 13:40:27 发布

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本文介绍了一种解决区间最大值查询问题的有效算法。通过预处理技术构造数据结构，实现快速查询给定区间内的最大值，适用于多种编程竞赛场景。

题目：

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

The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1862 Accepted Submission(s): 1475

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

Sample Output

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<ctype.h>    //tower()
#include<set>  
#include<map>  
#include<iomanip>// cout<<setprecision(1)<<fixed<<a;
#include<vector>   
#include<time.h>  
#include<assert.h>  //assert
#include<cmath>	
#include<algorithm>
#include<bitset>
#include<limits.h>
#include<stack>
#include<queue>
using namespace std;
const int maxn=1010;
const int inf=INT_MAX;

int t,n,q,a[maxn],sti[maxn][16];

void initrmq(){//记录最大值
	for(int i=0;i<n;++i) sti[i][0]=a[i];
	for(int j=1;(1<<j)<=n;++j){
		for(int i=0;i+(1<<j)-1<n;++i){
			sti[i][j]=max(sti[i][j-1],sti[i+(1<<(j-1))][j-1]);
		}
	}
}

int rmq(int u,int v){
	int k=(int)(log(v-u+1.0)/log(2.0));
	return max(sti[u][k],sti[v-(1<<k)+1][k]);
}

int main(){// hdu 5443  0MS	1644K
	int c,d;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=0;i<n;++i){
			scanf("%d",&a[i]);
		}
		initrmq();
		scanf("%d",&q);
		while(q--){
			scanf("%d%d",&c,&d);
			printf("%d\n",rmq(c-1,d-1));
		}
	}
	return 0;
}