hdu 2838 Cow Sorting（树状数组）-优快云博客

本文链接：https://blog.youkuaiyun.com/xuejye/article/details/75013173

本文介绍了一道关于重新排列一群奶牛以减少其‘暴躁值’总交换时间的算法问题CowSorting。通过使用树状数组来高效计算逆序对及其累积暴躁值之和，实现了最小化调整所需的时间。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2838点击打开链接

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3522 Accepted Submission(s): 1222

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

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：树状数组用两个数组分别储存逆序的个数和比nn大的数的和因为每次需要两个都移动因此逆序个数×nn花费固定加上比nn大的数的和结果就为所求

此题long long 过不了用__int 64

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <vector>
using namespace std;
__int64 tree[111111];
__int64 value[111111];
__int64 lowbit(__int64 x)
{
    return x&-x;
}
__int64 getsumtree(__int64 x)
{
    __int64 s=0;
    while(x>0)
    {
        s+=tree[x];
        x-=lowbit(x);
    }
    return s;
}
__int64 getsumvalue(__int64 x)
{
    __int64 s=0;
    while(x>0)
    {
        s+=value[x];
        x-=lowbit(x);
    }
    return s;
}
void add(__int64 x,__int64 num,__int64 val)
{
    while(x<=100000)
    {
        value[x]+=val;
        tree[x]+=num;
        x+=lowbit(x);
    }
}

int main()
{
   __int64 n=0;__int64 sum=0;
   scanf("%d",&n);
   for(__int64 i=1;i<=n;i++)
   {
       __int64 nn=0;
       scanf("%I64d",&nn);
       add(nn,1,nn);
       sum+=(getsumtree(n)-getsumtree(nn))*nn+(getsumvalue(n)-getsumvalue(nn));
   }
   printf("%I64d\n",sum);
}