hdu 2838 Cow Sorting（树状数组）

最新推荐文章于 2025-09-09 08:37:13 发布

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最新推荐文章于 2025-09-09 08:37:13 发布

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本文介绍了一个关于牛的排序问题，旨在通过仅交换相邻元素的方式将数组按升序排列，并计算出完成排序所需的最小代价。文章提供了一种算法实现思路，利用树状数组来高效地计算比当前元素大的前驱数量及其总和。

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2239 Accepted Submission(s): 711

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：给出一组数从1到N打乱，要求把数组又一次有序（从小到大），仅仅能交换相邻的两个数字。代价为相邻两个数字和。求最小代价？

思路：对于每一个数字x，我们仅仅须要把它和前面比它大的数字交换。求出交换代价，反复运行就能得出答案。

这个代价就是，比它大的数字个数t*x+前面比它大的数字和。

<pre name="code" class="cpp">#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<vector>
#include<functional>
#include<iostream>
using namespace std;
#define N 100005
#define ll __int64
int a[N],cnt[N],n;   //记录数字个数
ll sum[N];      //记录数字和
int lowbit(int x)
{
    return x&(-x);
}
void add(int x)
{
    int d=x;
    while(x<=n)
    {
        cnt[x]++;
        sum[x]+=d;
        x+=lowbit(x);
    }
}
int sum1(int x)   //求比x小的数字已经出现几个（包含x）
{
    int s=0;
    while(x)
    {
        s+=cnt[x];
        x-=lowbit(x);
    }
    return s;
}
ll sum2(int x)  //求当前出现的比x大的数字和
{
    ll s=0;
    while(x)
    {
        s+=sum[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int i,k,t;
    while(scanf("%d",&n)!=-1)
    {
        memset(cnt,0,sizeof(cnt));
        memset(sum,0,sizeof(sum));
        ll ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            add(a[i]);
            t=sum1(a[i]);  
            k=i-t;   //前面有几个比x大的数
            if(k!=0)
            {
                ans+=(ll)a[i]*k;       //注意数字会超出int
                ans+=sum2(n)-sum2(a[i]);  //求当前位置出现的比x大的数字和
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}