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

 1 #include<cstdio>
 2 #define F(i,a,b) for(int i=a;i<=b;i++)
 3 typedef long long LL;
 4 const int N=1e5+7;
 5 int num[N],n,x,cnt;
 6 LL sum[N],ans;    
 7 
 8 inline void add(int x,int k){for(;x<=100000;x+=x&-x)num[x]+=1,sum[x]+=k;}
 9 inline int asknum(int x){int an=0;while(x>0)an+=num[x],x-=x&-x;return an;}
10 inline LL asksum(int x){LL an=0;while(x>0)an+=sum[x],x-=x&-x;return an;}
11 
12 int main(){
13     while(~scanf("%d",&n)){
14         F(i,1,N-1)num[i]=0,sum[i]=0;ans=0;
15         F(i,1,n){
16             scanf("%d",&x),add(x,x),cnt=i-asknum(x);
17             if(cnt)ans+=(LL)cnt*x+asksum(100000)-asksum(x);
18         }
19         printf("%lld\n",ans);
20     }
21     return 0;
22 }

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转载于:https://www.cnblogs.com/bin-gege/p/5696110.html