hdoj Cow Sorting

標籤：樹狀數組

題意：給你n個無序的數，通過相鄰的兩個數交換，交換的價值是兩個數之和，求使之increasing order的最小代價和
與poj3270不同的是poj可以每個數任意交換，可以用置換群來做

#include<cstdio>#include<cstring>#include<vector>#include<iostream>#include<algorithm>using namespace std;#define ll __int64#define N 100000+10ll a[N],c[N],s[N],Sum[N];int n;ll lowbit(ll x){        return x&(-x);}void updata(ll t,ll value){        for(int i=t;i<=n;i+=lowbit(i))        {                c[i]+=value;                s[i]+=t;        }}int getSum(ll x,ll &y){        y=0;        ll temp=0;        for(int i=x;i>=1;i-=lowbit(i))        {                temp+=c[i];                y+=s[i];        }       return temp;}int main(){   while(scanf("%d",&n)!=EOF)   {           ll sum=0;           Sum[0]=0;           for(ll i=1;i<=n;i++)                   {                           scanf("%I64d",&a[i]);                           Sum[i]=Sum[i-1];                           Sum[i]+=a[i];                   }           for(ll i=1;i<=n;i++)           {                   updata(a[i],1);                   ll y;                   ll temp=i-getSum(a[i],y);                   sum+=temp*a[i]+Sum[i]-y;           }           printf("%I64d\n",sum);   }   return 0;}

hdoj Cow Sorting