HDU 2838 Cow Sorting【樹狀數組+逆序數】

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).

題意：給你一個n個數的序列，求逆序數對的和；

分析：用樹狀數組去維護動态求和的過程，主要是插入位置和排序後位置的來回切換；

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int MAXN =  + ;
LL n;

struct node {
    LL cnt;
    LL sum;
}s[MAXN];

LL lowbit(LL x) {
    return x & (-x);
}

inline void update(LL i, LL val, LL x) {
    while(i <= n) {
        s[i].cnt += x;
        s[i].sum += val;
        i += lowbit(i);
    }
}

inline LL query_cnt(LL x) {
    LL sum = ;
    while(x > ) {
        sum += s[x].cnt;
        x -= lowbit(x);
    }
    return sum;
}

inline LL query_sum(LL x) {
    LL ans = ;
    while(x > ) {
        ans += s[x].sum;
        x -= lowbit(x);
    }
    return ans;
}

int main() {
    while(~scanf("%lld", &n)) {
        memset(s, , sizeof(s));
        LL ans = , ant = , x;
        for(LL i = ; i <= n; ++i) {
            scanf("%lld", &x);
            update(x, x, L);
            LL tt1 = x - query_cnt(x);
            LL res1 = tt1 * x + (x * (x - ) / ) - query_sum(x - );
            ant += res1;
//          LL tt = i - query_cnt(x); //兩種寫法都可以 
//          LL res = tt * x + query_sum(n) - query_sum(x);
//          ans += res;
        }
        printf("%lld\n", ant);
    }
    return ;
}