zoukankan      html  css  js  c++  java
  • POJ 2750 Potted Flower

    Potted Flower
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 3872   Accepted: 1446

    Description

    The little cat takes over the management of a new park. There is a large circular statue in the center of the park, surrounded by N pots of flowers. Each potted flower will be assigned to an integer number (possibly negative) denoting how attractive it is. See the following graph as an example:

    (Positions of potted flowers are assigned to index numbers in the range of 1 ... N. The i-th pot and the (i + 1)-th pot are consecutive for any given i (1 <= i < N), and 1st pot is next to N-th pot in addition.)



    The board chairman informed the little cat to construct "ONE arc-style cane-chair" for tourists having a rest, and the sum of attractive values of the flowers beside the cane-chair should be as large as possible. You should notice that a cane-chair cannot be a total circle, so the number of flowers beside the cane-chair may be 1, 2, ..., N - 1, but cannot be N. In the above example, if we construct a cane-chair in the position of that red-dashed-arc, we will have the sum of 3+(-2)+1+2=4, which is the largest among all possible constructions.

    Unluckily, some booted cats always make trouble for the little cat, by changing some potted flowers to others. The intelligence agency of little cat has caught up all the M instruments of booted cats' action. Each instrument is in the form of "A B", which means changing the A-th potted flowered with a new one whose attractive value equals to B. You have to report the new "maximal sum" after each instruction.

    Input

    There will be a single test data in the input. You are given an integer N (4 <= N <= 100000) in the first input line.

    The second line contains N integers, which are the initial attractive value of each potted flower. The i-th number is for the potted flower on the i-th position.

    A single integer M (4 <= M <= 100000) in the third input line, and the following M lines each contains an instruction "A B" in the form described above.

    Restriction: All the attractive values are within [-1000, 1000]. We guarantee the maximal sum will be always a positive integer.

    Output

    For each instruction, output a single line with the maximum sum of attractive values for the optimum cane-chair.

    Sample Input

    5
    3 -2 1 2 -5
    4
    2 -2
    5 -5
    2 -4
    5 -1
    

    Sample Output

    4
    4
    3
    5
    

    Source

    POJ Monthly--2006.01.22,Zeyuan Zhu
      
     线段树+dp 感觉好经典啊  log(n)就能求出一串数字的最大的连续和
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #define N 100010
    using namespace std;
    struct num
    {
        int l,r,sum,maxsum,minsum,maxl,maxr,minl,minr;
    }a[4*N];
    int b[N];
    int main()
    {
        //freopen("data.in","r",stdin);
        void init(int k,int l,int r);
        void update(int k,int pos,int val);
        int n;
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&b[i]);
            }
            init(1,1,n);
            int m;
            scanf("%d",&m);
            while(m--)
            {
                int x,y;
                scanf("%d %d",&x,&y);
                update(1,x,y);
                if(a[1].sum==a[1].maxsum)
                {
                    printf("%d
    ",a[1].sum-a[1].minsum);
                }else
                {
                    printf("%d
    ",max(a[1].maxsum,a[1].sum-a[1].minsum));
                }
            }
        }
        return 0;
    }
    void pushup(int k)
    {
        int left=k<<1,right=k<<1|1;
        a[k].sum=a[left].sum+a[right].sum;
        a[k].maxsum=max(max(a[left].maxsum,a[right].maxsum),a[left].maxr+a[right].maxl);
        a[k].minsum=min(min(a[left].minsum,a[right].minsum),a[left].minr+a[right].minl);
        a[k].maxl=max(a[left].maxl,a[left].sum+a[right].maxl);
        a[k].maxr=max(a[right].maxr,a[right].sum+a[left].maxr);
        a[k].minl=min(a[left].minl,a[left].sum+a[right].minl);
        a[k].minr=min(a[right].minr,a[right].sum+a[left].minr);
    }
    void init(int k,int l,int r)
    {
        a[k].l=l; a[k].r=r;
        if(l==r)
        {
            a[k].sum=a[k].maxsum=a[k].minsum=a[k].maxr=a[k].maxl=a[k].minr=a[k].minl=b[l];
            return ;
        }
        int mid=(l+r)>>1;
        init(k<<1,l,mid);
        init(k<<1|1,mid+1,r);
        pushup(k);
    }
    void update(int k,int pos,int val)
    {
        if(a[k].l==a[k].r)
        {
            a[k].sum=a[k].maxsum=a[k].minsum=a[k].maxr=a[k].maxl=a[k].minr=a[k].minl=val;
            return ;
        }
        int mid=(a[k].l+a[k].r)>>1;
        if(mid>=pos)
        {
            update(k<<1,pos,val);
        }else
        {
            update(k<<1|1,pos,val);
        }
        pushup(k);
    }
    


  • 相关阅读:
    Http协议的断点续传下载器,使用观察者模式监视下载进度,使用xml保存下载进度。
    C++ 复制到粘贴板
    编译防火墙——C++的Pimpl惯用法解析
    字符串输出
    windows路径操作API函数
    Boost解析xml——xml写入
    智能指针shared_ptr
    Boost 解析xml——插入Item
    ListCtrl添加右键菜单(在对话框类中)
    抓包工具Charles的使用说明
  • 原文地址:https://www.cnblogs.com/suncoolcat/p/3295165.html
Copyright © 2011-2022 走看看