树状数组求逆序对 附HDU1394

清楚的图解，解释树状数组求逆序对的方法，转载：https://blog.csdn.net/ssimple_y/article/details/53744096

题目：https://vjudge.net/problem/HDU-1394

思路：因为数字[1,n]，把a[i]放到末尾，逆序对数量会增加n-a[i]个，即比它大的数字个数，

减少a[i]-1个，即比它小的数字个数

#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <string>
#include <stack>
#include <vector>
#include <list>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 5e3+10;
int a[N],c[N];
int n;

inline int lb(int x){
    return x&(-x);
}

void update(int inx){
    for(int i = inx; i <= n; i += lb(i)){
        ++c[i];
    }
}

int sum(int inx){
    int res = 0;
    for(int i = inx; i >= 1; i -= lb(i)){
        res += c[i]; 
    }

    return res;
}

void solve(){

    while(~scanf("%d",&n)){
        memset(c,0,sizeof(c));
        int tot = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d",a+i);
            update(++a[i]);  //这里处理一下0，使得更新和查询不会死循环
            tot += i - sum(a[i]);
        }
        int Min = tot;
        for(int i = 1; i <= n; ++i){
            tot += n - 2*a[i] + 1;
            Min = min(tot,Min);
        }
        printf("%d
",Min);
    }

}

int main(){
 
  //  ios::sync_with_stdio(false);
  //  cin.tie(0); cout.tie(0);
    solve();

    return 0;
}