An Easy Problem

最新推荐文章于 2021-06-26 11:54:32 发布

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#单调DP

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本文介绍了一种使用单调动态规划(monotonic dynamic programming, DP)和栈(stack)解决特定查询问题的方法。通过定义一个特殊函数f(j)，并在给定数组a的情况下进行查询，实现了高效的查询响应。文章还提供了一个具体的实现示例，包括输入输出格式、样例及其解释。

Description

Now we define that:

A(j)={i|0<i<j&&a[i]<a[j]}

Here a is an array with n element(1<=n<=1000000);

And then we define a function f that:

int f(int j){

if(A(j) is not empty) return max( A(j) );

else return -1;

}

Input

The first line contain an integer n(1<=n<=1000000),the size of the array a.

The second line contain n integers describe the array a.

The third line is an integer m(1<=m<=1000000),the number of the query.

And then follow m lines,each line is an integer j(1<=j<=n).

Output

For each query j,output a[f(j)] if f(j)>=1,Otherwise output -1 instead.

Sample Input

1 2 3

Sample Output

-1

单调DP，用另一数组记录下a[f(i)]，可用stack实现。

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
#define N 1000005
int a[N],b[N];
int main()
{
	int n,m,i,temp,x;
	stack<int> S;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		if(S.empty())
		{
			b[i]=-1;
			S.push(a[i]);
		}
		else
		{
			temp=S.top();
			if(temp<a[i])
			{
				b[i]=temp;
				S.push(a[i]);
			}
			else
			{
				S.pop();
				while(!S.empty())
				{
					temp=S.top();
					if(temp<a[i])
					{
						b[i]=temp;
						S.push(a[i]);
						break;
					}
					else
						S.pop();
				}
				if(S.empty())
				{
					b[i]=-1;
					S.push(a[i]);
				}
			}
		}
	}
	scanf("%d",&m);
	while(m--)
	{
		scanf("%d",&x);
		printf("%d\n",b[x-1]);
	}
	return 0;
}