HDU 4911 Inversion 树状数组求逆序数对

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显然每次交换都能减少1

所以求出逆序数对数，然后-=k就好了。。。

_(:зゝ∠)_

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<set>#include<map>#include<iostream>#include<algorithm>using namespace std;#define N 100005#define ll long longll c[N+100000], maxn;  inline ll Lowbit(ll x){return x&(-x);}  void change(ll i, ll x)//i点增量为x  {      while(i <= maxn)      {          c[i] += x;          i += Lowbit(i);      }  }ll sum(ll x){//区间求和 [1,x]      ll ans = 0;      for(ll i = x; i >= 1; i -= Lowbit(i))          ans += c[i];      return ans;  }ll a[N], n, k;set<ll>s;set<ll>::iterator p;map<ll,ll>mp;int main(){    ll i;    while(cin>>n>>k){        s.clear(); mp.clear();        for(i = 1; i <= n; i++)scanf("%I64d",&a[i]), s.insert(a[i]);        maxn = n+100;        for(p = s.begin(), i = 2; p!=s.end(); p++, i++)        {            mp[*p] = i;        }        for(i = 1; i <= n; i++)a[i] = mp[a[i]];        memset(c, 0, sizeof c);        ll ans = 0;        for(i = n; i >= 1; i--)        {            ans += sum(a[i]-1);            change(a[i], 1);        }        ans -= k;        cout<< max(0ll, ans) <<endl;    }    return 0;}

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