codeforces 487B B. Strip(RMQ+二分+dp)

最新推荐文章于 2024-08-28 22:49:31 发布

转载最新推荐文章于 2024-08-28 22:49:31 发布 · 115 阅读

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原文链接：http://www.cnblogs.com/zhangchengc919/p/5832963.html

本文介绍了一种将一串数字按特定条件进行最优分割的方法。使用动态规划与范围查询技术来确定最少分割次数，确保每个部分至少包含l个数字且最大最小值之差不超过s。通过RMQ预处理加速区间查询。

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题目链接：

B. Strip

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

input

7 2 2
1 100 1 100 1 100 1

output

-1

题意：

把n个数字分成尽量少的连续部分,每部分至少l个数,而且最大值和最小值得差不超过s,问最少能分成多少部分;

思路：

dp[i]表示前i个数字最少能分成多少部分,因为差不能超过s，所以先求出RMQ,以方便后面询问,可以二分找出区间[l,i]l是满足要求的最左边的第一个差值不超过s的位置,然后看是否满足长度的要求,还有就是要转移最小的值;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}
 
const LL mod=1e6+3;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=1e3+520;
const double eps=1e-12;

int n,s,le,a[N],dp[N];
int mi[N][20],ma[N][20];
inline void RMQ()
{
    for(int i=0;i<n;i++)ma[i][0]=mi[i][0]=a[i];
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=0;i+(1<<j)<=n;i++)
            {
                mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
                ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);
            }
    }
}
int query(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=R-L+1)k++;
    return max(ma[L][k],ma[R-(1<<k)+1][k])-min(mi[L][k],mi[R-(1<<k)+1][k]);
}

int check(int x,int y)
{
    int temp=query(x,y);
    if(temp<=s)return 1;
    return 0;
}
int main()
{
    read(n);read(s);read(le);
    For(i,0,n-1)read(a[i]),dp[i]=inf;
    RMQ();
    For(i,0,n-1)
    {
        int l=0,r=i;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(check(mid,i))r=mid-1;
            else l=mid+1;
        }
        r++;
        
        if(i-r+1<le)dp[i]=inf;
        else 
        {
            if(r==0)dp[i]=1;
            else 
            {
                for(int j=r-1;j<=i-le;j++)dp[i]=min(dp[i],dp[j]+1);
            }
        }
    }
    if(dp[n-1]==inf)dp[n-1]=-1;
    cout<<dp[n-1]<<endl;
    return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5832963.html