Codeforces Round #333 (Div. 2) B. Approximating a Constant Range (线段树区间最值)

B. Approximating a Constant Range

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium
intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest
problem while we're at it?
You're given a sequence of n data
points a1, ..., an.
There aren't any big jumps between consecutive data points — for each 1 ≤ i < n,
it's guaranteed that |ai + 1 - ai| ≤ 1.
A range [l, r] of
data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1.
Formally, let M be the maximum and m the
minimum value of ai for l ≤ i ≤ r;
the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.

Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) —
the number of data points.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output
Print a single number — the maximum length of an almost constant range of the given sequence.

Sample test(s)

input

5
1 2 3 3 2

output

4

input

11
5 4 5 5 6 7 8 8 8 7 6

output

5

Note
In the first sample, the longest almost constant range is [2, 5];
its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10];
the only almost constant range of the maximum length 5 is [6, 10].

题意：给你一个序列，找出最长的区间，这个区间的最大值和最小值相差1
思路：i,j两个指针j指向开头，如果一个区间的最大值和最小值相差大于1，那么它所属的区间也一定不满足条件，那么就不必计算，那么j的指针向前进，然后外面的i也不断前进
总结：刚开始觉得线段树写肯定不行，然后开始找规律，并没有找到，就随便写了一个普通的区间线段树交了，TLE了，然后开始在线段树基础上找规律，列了几组数据找规律，最后终于找到。。还是要有耐心。。

ac代码：
#include<stdio.h>
#define MAXN 100100
int MIN(int a,int b)
{
return a>b?b:a;
}
int MAX(int a,int b)
{
return a>b?a:b;
}
struct s
{
int left;
int right;
int min;
int max;
}tree[MAXN*3];
int num[MAXN];
void build(int l,int r,int i)
{
tree[i].left=l;
tree[i].right=r;
if(l==r)
{
tree[i].min=num[l];
tree[i].max=num[l];
}
else
{
int mid;
mid=(l+r)/2;
build(l,mid,i*2);
build(mid+1,r,i*2+1);
tree[i].min=MIN(tree[i*2].min,tree[i*2+1].min);
tree[i].max=MAX(tree[i*2].max,tree[i*2+1].max);
}
}
int query1(int i,int l,int r)
{
if(tree[i].left==l&&tree[i].right==r)
{
return tree[i].min;
}
if(r<=tree[i*2].right)
{
return query1(i*2,l,r);
}
if(l>=tree[i*2+1].left)
{
return query1(i*2+1,l,r);
}
int mid=(tree[i].left+tree[i].right)/2;
return MIN(query1(i*2,l,mid),query1(i*2+1,mid+1,r));
}
int query2(int i,int l,int r)
{
if(tree[i].left==l&&tree[i].right==r)
{
return tree[i].max;
}
if(r<=tree[i*2].right)
{
return query2(i*2,l,r);
}
if(l>=tree[i*2+1].left)
{
return query2(i*2+1,l,r);
}
int mid=(tree[i].left+tree[i].right)/2;
return MAX(query2(i*2,l,mid),query2(i*2+1,mid+1,r));
}
int main()
{
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
build(1,n,1);
int ans;
int j=1;
for(i=1;i<=n;i++)
{
int M=query2(1,j,i);
int mi=query1(1,j,i);
while(M-mi>1&&j<=i)
{
j++;
M=query2(1,j,i);
mi=query1(1,j,i);
}
ans=MAX(ans,i-j+1);
}
printf("%d\n",ans);
}
return 0;
}