hdu 2838求逆序數開了兩個數組

http://acm.hdu.edu.cn/showproblem.php?pid=2838

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4387    Accepted Submission(s): 1580

 

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).

 

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

 

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 題意：和以前求逆序數的題意一樣，不過這不是求總共多少個逆序數，而是求如a[i-1]>a[i]花費就是a[i-1]+a[i]，問使整個序列上升有序最小的花費。以前是求得使整個序列上升有序最小的交換次數。這個也一樣。

//#include<bits/stdc++.h>
//using namespace std;
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=1000010;
ll n,c[3][maxn];
ll lowbit(ll x)
{
    return x&(-x);
}
ll query(ll flag,ll x)
{
    ll res=0;
    while(x)
    {
        res+=c[flag][x];
        x-=lowbit(x);
    }
    return res;
}
void add(ll flag,ll x,ll val)
{
    while(x<=n)
    {
        c[flag][x]+=val;
        x+=lowbit(x);
    }
}
ll a[maxn];
int main()
{
    ll tmp1,tmp2,ans,i;
    while(scanf("%lld",&n)==1&&n)
    {
        ll ans=0;
        memset(c,0,sizeof(c));
        for(i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        for(i=n;i>=1;i--)
        {
            add(1,a[i],1);
            add(2,a[i],a[i]);
            tmp1=query(1,a[i]-1);
            tmp2=query(2,a[i]-1);
            ans+=tmp1*a[i]+tmp2;
        }
        printf("%lld\n",ans);
    }
}