• hdu 2838 Cow Sorting 树状数组求所有比x小的数的个数


    Cow Sorting

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4766    Accepted Submission(s): 1727


    Problem Description
    Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.

    Please help Sherlock calculate the minimal time required to reorder the cows.
     
    Input
    Line 1: A single integer: N
    Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.
     
    Output
    Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.
     
    Sample Input
    3 2 3 1
     
    Sample Output
    7
    Hint
    Input Details Three cows are standing in line with respective grumpiness levels 2, 3, and 1. Output Details 2 3 1 : Initial order. 2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4). 1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).
     
     
    题意: n个数的排列,每次可以互换相邻的元素,最终变成一个递增的序列,每次互换的代价为互换的两个数的和,求最小代价。
     
    题解:从无序到递增过程,就是冒泡排序的过程。对于每一个元素a[i],需要交换的次数就是在a[1]~a[i-1]中,所有比a[i]大的元素的个数cnt,a[i]到最终状态花费的代价就是cnt*a[i]+所有被交换的元素的和sum
    树状数组处理出  比a[i]小的数的个数b[i] 和 所有比a[i]小的数的和c[i]就可以直接处理
     
    #include<iostream>
    #include<string.h>
    #define ll long long
    using namespace std;
    ll a[100005],b[100005],c[100005];
    //a[i]保存原始数据,b[i]保存比a[i]小的数的个数,c[i]保存所有比a[i]小的数的和
    ll lowbit(ll x)
    {
        return x&(-x);
    }
    
    ll getnum(ll x)//求比x小的数的个数
    {
        ll cnt=0;
        while(x>0)
        {
            cnt=cnt+b[x];
            x=x-lowbit(x);
        }
        return cnt;
    }
    
    ll getsum(ll x)//求比x小的数的和
    {
        ll ans=0;
        while(x>0)
        {
            ans=ans+c[x];
            x=x-lowbit(x);
        }
        return ans;
    }
    
    void add(ll x,ll y)//更新,对第x个位置的数进行更新,y是更新值
    {
        while(x<=100000)
        {
            b[x]=b[x]+1;
            c[x]=c[x]+y;
            x=x+lowbit(x);
        }
    }
    int main()
    {
        int n;
        while(~scanf("%d",&n))
        {
            ll ans=0,sum=0;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            for(int i=1;i<=n;i++)
            {
                scanf("%lld",&a[i]);
                add(a[i],a[i]);//将第x个位置的值,修改为x
                sum=sum+a[i];//sum求所有数的和
                ans=ans+a[i]*(i-getnum(a[i]));//i-getnum(a[i])是比a[i]大的数的个数
                ans=ans+sum-getsum(a[i]);//sum-getsum(a[i])是所有比a[i]大的数的和
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }
  • 相关阅读:
    为什么 JVM 不用 JIT 全程编译?
    JVM Internals
    JIT与JVM的三种执行模式:解释模式、编译模式、混合模式
    Dart编译技术与平台
    Dart 库预览
    使用VSCode开发Flutter
    环境变量
    使用Homebrew管理你的mac开发包
    brew 又叫Homebrew,是Mac OSX上的软件包管理工具
    使用async/await消除callback hell
  • 原文地址:https://www.cnblogs.com/-citywall123/p/11559424.html
Copyright © 2020-2023  润新知