POJ 2452 Sticks Problem(二分+RMQ)

题意：给定一串数列，求满足ai<=ak<=aj条件（ak不一定是有序的）的最大j-i，即端点为这段序列的最小值和最大值。

思路：二分+RMQ，可以这样做，枚举i，求出比a[i]小的第一个数字的下标r，那么在[i,r]范围内找到最大值的下标k，那么就可以更新答案了。那么为什么在求比a[i]小的第一个数字的时候可以用二分呢？因为rmq数组其实满足单调性的，所以可以二分

吐槽：一开始没看清题目就做..以为是求LIS...无语..

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 50000+100;

int dmax[maxn][20];
int dmin[maxn][20];
int d[maxn];
int mmax(int i,int j)
{
	if (d[i]>d[j])
		return i;
	return j;
}
int mmin(int i,int j)
{
	if (d[i]<d[j])
		return i;
	return j;
}
void initmax(int n)
{
	for (int i = 1;i<=n;i++)
		dmax[i][0]=i;
	for (int j = 1;(1<<j)<=n;j++)
		for (int i = 1;i+(1<<j)-1<=n;i++)
			dmax[i][j]=mmax(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);
}
int getmax(int L,int R)
{
	int k = 0;
	while ((1<<(k+1)) <= R-L+1)
		k++;
	return mmax(dmax[L][k],dmax[R-(1<<k)+1][k]);
}

void initmin(int n)
{
	for (int i = 1;i<=n;i++)
		dmin[i][0]=i;
	for (int j = 1;(1<<j)<=n;j++)
		for (int i = 1;i+(1<<j)-1<=n;i++)
			dmin[i][j]=mmin(dmin[i][j-1],dmin[i+(1<<(j-1))][j-1]);
}
int getmin(int L,int R)
{
	int k = 0;
	while ((1<<(k+1)) <= R-L+1)
		k++;
	return mmin(dmin[L][k],dmin[R-(1<<k)+1][k]);
}
int get_r(int i,int l,int r)
{
    while (l<r)
	{
		int m = (l+r)/2;
		if (d[i]<d[getmin(l,m)])
			l=m+1;
		else
			r=m;
	}
	return l;
}
int main()
{
	int n,q;
	while (scanf("%d",&n)!=EOF)
	{
		for (int i = 1;i<=n;i++)
			scanf("%d",&d[i]);
		initmax(n);
		initmin(n);
		int ans = 0;
		for (int i = 1;i<=n;i++)
		{
			int r = get_r(i,i+1,n);
			int k = getmax(i,r);
			if (d[i]<d[k])
			  ans = max(ans,k-i);
		}
		if (!ans)
			printf("-1\n");
		else
			printf("%d\n",ans);
	}
}

Description

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by S1, S2, S3, ...Sn. After measuring the length of each stick Sk (1 <= k <= n), she finds that for some sticks Si and Sj (1<= i < j <= n), each stick placed between Si and Sj is longer than Si but shorter than Sj. 

Now given the length of S1, S2, S3, …Sn, you are required to find the maximum value j - i.

Input

The input contains multiple test cases. Each case contains two lines. 
Line 1: a single integer n (n <= 50000), indicating the number of sticks. 
Line 2: n different positive integers (not larger than 100000), indicating the length of each stick in order.

Output

Output the maximum value j - i in a single line. If there is no such i and j, just output -1.

Sample Input

4
5 4 3 6
4
6 5 4 3

Sample Output

1
-1