zoukankan      html  css  js  c++  java
  • Codeforces Round #278 (Div. 2) D. Strip 线段树优化dp

    D. 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.

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    #define pi (4*atan(1.0))
    #define eps 1e-14
    #define bug(x)  cout<<"bug"<<x<<endl;
    const int N=1e5+10,M=1e6+10,inf=2e9;
    const ll INF=1e18+10,mod=2147493647;
    int a[N];
    struct linetree
    {
        int maxx[N<<2],minn[N<<2];
        void pushup(int pos)
        {
            maxx[pos]=max(maxx[pos<<1],maxx[pos<<1|1]);
            minn[pos]=min(minn[pos<<1],minn[pos<<1|1]);
        }
        void build(int l,int r,int pos)
        {
            if(l==r)
            {
                maxx[pos]=a[l];
                minn[pos]=a[l];
                return;
            }
            int mid=(l+r)>>1;
            build(l,mid,pos<<1);
            build(mid+1,r,pos<<1|1);
            pushup(pos);
        }
        void update(int p,int c,int l,int r,int pos)
        {
            if(l==r)
            {
                maxx[pos]=c;
                minn[pos]=c;
                return;
            }
            int mid=(l+r)>>1;
            if(p<=mid)
                update(p,c,l,mid,pos<<1);
            else
                update(p,c,mid+1,r,pos<<1|1);
            pushup(pos);
        }
        int query(int L,int R,int l,int r,int pos,int flag)
        {
            if(L<=l&&r<=R)
            {
                if(flag)
                    return maxx[pos];
                else
                    return minn[pos];
            }
            int mid=(l+r)>>1;
            int ans=-inf;
            if(!flag)
                ans=inf;
            if(L<=mid)
                if(flag)
                    ans=max(ans,query(L,R,l,mid,pos<<1,flag));
                else
                    ans=min(ans,query(L,R,l,mid,pos<<1,flag));
            if(R>mid)
                if(flag)
                    ans=max(ans,query(L,R,mid+1,r,pos<<1|1,flag));
                else
                    ans=min(ans,query(L,R,mid+1,r,pos<<1|1,flag));
            return ans;
        }
    };
    linetree tree,dp;
    int main()
    {
        int n,l,s;
        scanf("%d%d%d",&n,&s,&l);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        tree.build(1,n+1,1);
        dp.build(1,n+1,1);
        dp.update(1,0,1,n+1,1);
        for(int i=1;i<=n;i++)
        {
            int st=0,en=i-l,ans=-1;
            while(st<=en)
            {
                int mid=(st+en)>>1;
                if(tree.query(mid+1,i,1,n+1,1,1)-tree.query(mid+1,i,1,n+1,1,0)<=s)
                {
                    ans=mid;
                    en=mid-1;
                }
                else
                    st=mid+1;
            }
            //cout<<ans<<endl;
            if(ans==-1||ans+1>i-l+1)
                dp.update(i+1,inf,1,n+1,1);
            else
            {
                int minn=dp.query(ans+1,i-l+1,1,n+1,1,0);
                //cout<<ans+1<<" "<<i-l+1<<" "<<i<<" "<<minn<<endl;
                dp.update(i+1,minn+1,1,n+1,1);
            }
    
        }
        if(dp.query(n+1,n+1,1,n+1,1,1)>=inf)
            printf("-1");
        else
            printf("%d
    ",dp.query(n+1,n+1,1,n+1,1,1));
        return 0;
    }
    ///  dp[i]=min(dp[mid-1]-dp[i-l])+1
  • 相关阅读:
    分部视图在ASP.NET MVC中的应用
    SVN、Git设置提交时忽略的文件
    TortoiseSVN使用步骤和trunk,Branch,Tag详细说明
    SQL Server系统表介绍与使用
    C#中Task的使用简单总结
    Auto Encoder用于异常检测
    mac bash 下使用vi 快捷方式——因为没有alt键 所以没有办法 用vi模式也非常方便的
    daal utils printNumericTable
    https 不会被中间人攻击——因为中间人即使拿到了数据,也是加密的
    Let's Encrypt 免费通配符 SSL 证书申请教程——但是也需要email,域名所有权等,如果是黑产用的话会这样用吗?会不会暴露自己身份???
  • 原文地址:https://www.cnblogs.com/jhz033/p/6491440.html
Copyright © 2011-2022 走看看