Codeforces 488D Strip【dp+RMQ--------ST】

B. Strip

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Alexandra has a paper strip with n numbers on it. Let's call them
ai from left to right.

Now Alexandra wants to split it into some pieces (possibly
1). For each piece of strip, it must satisfy:

Each piece should contain at least l numbers.
The difference between the maximal and the minimal number on the piece should be at most
s.
Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input
The first line contains three space-separated integers
n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105).

The second line contains n integers
ai separated by spaces ( - 109 ≤ ai ≤ 109).

Output
Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Examples

Input

7 2 2
1 3 1 2 4 1 2

Output

3

Input

7 2 2
1 100 1 100 1 100 1

Output

-1

Note
For the first sample, we can split the strip into 3 pieces:
[1, 3, 1], [2, 4], [1, 2].

For the second sample, we can't let 1 and
100 be on the same piece, so no solution exists.

题目大意：

给你一个长度为N的序列，我们将其分成若干个子序列（必须连续），需要保证每个子序列中的最大值和最小值的差不超过S，并且子序列中的元素的个数不能少于l，问最少可以分成几个子序列出来，如果不可能，输出-1.

思路：

1、对应查询区间最大值和最小值，那么我们使用ST算法是最好不过的，预处理很快，查询O（1）；

2、那么考虑dp，设定dp【i】表示前i个（包括第i个在内），最少能够分成几段。

那么不难推出其状态转移方程：

dp【i】=min（dp【j】+1，dp【i】），其中j是第一个合法的位子，使得从j+1到i保证有至少l个元素，且其区间值极差小于等于s、

Ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<map>
using namespace std;
int a[200005];
int maxn[200005][100];
int minn[200005][100];
int dp[200005];
int n,s,l;
void ST()
{
int len=floor(log10(double(n))/log10(double(2)));
for(int j=1;j<=len;j++)
{
for(int i=1;i<=n+1-(1<<j);i++)
{
maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
}
}
}
int query(int a,int b)
{
int len= floor(log10(double(b-a+1))/log10(double(2)));
return max(maxn[a][len], maxn[b-(1<<len)+1][len])-min(minn[a][len], minn[b-(1<<len)+1][len]);
}
int main()
{
while(~scanf("%d%d%d",&n,&s,&l))
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
maxn[i][0]=minn[i][0]=a[i];
}
ST();
memset(dp,0x3f3f3f3f,sizeof(dp));
int pre=0;
dp[0]=0;
for(int i=1;i<=n;i++)
{
while(i-pre>=l&&query(pre+1,i)>s||dp[pre]==0x3f3f3f3f)pre++;
if(i-pre>=l)dp[i]=min(dp[pre]+1,dp[i]);
}
if(dp
==0x3f3f3f3f)printf("-1\n");
else printf("%d\n",dp
);
}
}