codeforces-487B Strip（dp+rmq+二分+水数据）

笑死，最坏情况下时间复杂度为O(n*nlogn)

思路：每次找前面最小l，使[l,i]满足条件，dp[i] = dp[l-1]+1;

#include <stdio.h>
#include <string.h>
#include <ctime>
#include <stack>
#include <string>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lowbit(x) (x&-x)

const int maxn = 1e5+7;
const int INF = 0x3f3f3f3f;

int rmq[maxn][20][2];
int mm[maxn],a[maxn];
int dp[maxn],n,s;

void initRMQ()
{
    mm[0] = -1;
    for(int i=1; i<=n; i++)
    {
        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
        rmq[i][0][0] = rmq[i][0][1] = a[i];
    }
    for(int j=1; j<=mm[n]; j++)
        for(int i=1; i+(1<<j)-1<=n; i++)
    {
        rmq[i][j][0] = max(rmq[i][j-1][0],rmq[i+(1<<(j-1))][j-1][0]);
        rmq[i][j][1] = min(rmq[i][j-1][1],rmq[i+(1<<(j-1))][j-1][1]);
    }
}

int qMin(int x,int y)
{
    int k = mm[y-x+1];
    return min(rmq[x][k][1],rmq[y-(1<<k)+1][k][1]);
}

int qMax(int x,int y)
{
    int k = mm[y-x+1];
    return max(rmq[x][k][0],rmq[y-(1<<k)+1][k][0]);
}

int f(int l,int r)
{
    return qMax(l,r)-qMin(l,r);
}

int findLeft(int l,int r,int i)
{
    int m,ret = -1;
    while(l<=r)
    {
        m = (l+r)>>1;
        if(f(m,i)<=s)
        {
            ret = m;
            r = m-1;
        }
        else
            l = m+1;
    }
    return ret;
}

int main()
{
    int l;
    while(scanf("%d%d%d",&n,&s,&l)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        initRMQ();
        if(qMax(1,1+l-1)-qMin(1,1+l-1)>s)
        {
            //puts("this");
            puts("-1");
            continue;
        }
        for(int i=1; i<l; i++)
            dp[i] = maxn;
        dp[l] = 1;
        for(int i=l+1; i<=n; i++)
        {
            int t = findLeft(1,i-l+1,i);
            if(t!=-1)
            {
                while(dp[t-1]>=maxn&&t<=i-l)
                    t++;
                dp[i] = dp[t-1]+1;
            }
            else
                dp[i] = maxn;
        }

        if(dp[n]<maxn)
            printf("%d\n",dp[n]);
        else
            puts("-1");
    }
	return 0;
}

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

codeforces-487B Strip（dp+rmq+二分+水数据）

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