HDU 4911 Inversion 树状数组求逆序数对

显然每次交换都能降低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 long
ll 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;
}