hdu_2838_Cow Sorting(树状数组求逆序对)

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

题意：给你一串数，让你排序，只能交换相邻的数，每次交换花费交换的两个树的和，问最小交换的价值

题解：实质就是求逆序对

#include<cstdio>#define F(i,a,b) for(int i=a;i<=b;i++)typedef long long LL;const int N=1e5+7;int num[N],n,x,cnt;LL sum[N],ans;inline void add(int x,int k){for(;x<=100000;x+=x&-x)num[x]+=1,sum[x]+=k;}inline int asknum(int x){int an=0;while(x>0)an+=num[x],x-=x&-x;return an;}inline LL asksum(int x){LL an=0;while(x>0)an+=sum[x],x-=x&-x;return an;}int main(){while(~scanf("%d",&n)){F(i,1,N-1)num[i]=0,sum[i]=0;ans=0;F(i,1,n){scanf("%d",&x),add(x,x),cnt=i-asknum(x);if(cnt)ans+=(LL)cnt*x+asksum(100000)-asksum(x);}printf("%lld\n",ans);}return 0;}