HDU 1394 Minimum Inversion Number 树状数组

#include <map>
#include <set>
#include <list>
#include <cmath>
#include<cctype>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
#define MAXN 5005
int gcd(int a, int b)
{
return a % b == 0 ? b : gcd(b, a % b);
}
int C[MAXN];
int D[MAXN];
inline void insert(int pos)//注意这里是利用杂凑以下标与对应值判断元素个数，方便计算最后结果
{
while (pos<=MAXN)
{
C[pos]++;
pos+=(pos&(-pos));
}
}
inline int sum(int x)//这里并不是直接求元素的和，而是利用下标对应关系求的是元素个数的和
{
int sum=0;
while (x>0)
{
sum+=C[x];
x-=(x&(-x));
}
return sum;
}
int main()
{
int N;
while (scanf("%d",&N)==1)
{
memset(C,0,sizeof(C));
memset(D,0,sizeof(D));
int Sum=0,ans;int a;
for (int i=1;i<=N;i++)
{
scanf("%d",&D[i]);
D[i]++;
Sum+=(i-sum(D[i])-1);
insert(D[i]);
}
// printf("%d\n",Sum);
ans=Sum;int res=Sum;
for (int i=1;i<=N;i++)
{
Sum=Sum-(D[i]-1)+N-D[i];
ans=min(ans,Sum);
}
printf("%d\n",ans);
}
return 0;
}