HDU 4911 Inversion（求逆序对）

bobo has a sequence a 1,a 2,…,a n. He is allowed to swap two adjacent numbers for no more than k times. 

Find the minimum number of inversions after his swaps. 

Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a i>aj.

InputThe input consists of several tests. For each tests: 

The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).
OutputFor each tests: 

A single integer denotes the minimum number of inversions.
Sample Input

3 1
2 2 1
3 0
2 2 1

Sample Output

1
2

题解：

题意：

给你n个数，允许相邻的交换k次，问你能得到的最小逆序对数

思路：

一开始想用树状数组来求逆序对。。后来发现1e9就放弃了。。老老实实归并排序吧，max(逆序对数减去交换次数,0)就是答案，注意要long long

代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
using namespace std;
#define INF 100861111
#define ll long long
#define eps 1e-15
int a[100005];
int t[100005];
ll ans;
void Merge(int l,int r,int mid)
{
int i=l,j=mid+1,k=l;
while(i<=mid&&j<=r)
{
if(a[i]>a[j])
{
ans+=(mid+1-i);
t[k]=a[j];
j++;
}
else
{
t[k]=a[i];
i++;
}
k++;
}
while(i<=mid)
{
t[k]=a[i];
i++;
k++;
}
while(j<=r)
{
t[k]=a[j];
k++;
j++;
}
for(i=l;i<=r;i++)
a[i]=t[i];
}
void guibin(int l,int r)
{
if(l<r)
{
int mid=(l+r)/2;
guibin(l,mid);
guibin(mid+1,r);
Merge(l,r,mid);
}
}
int main()
{
int i,j,n,x;
ll k;
while(scanf("%d%lld",&n,&k)!=EOF)
{
ans=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
guibin(0,n-1);
ans=max(ans-k,(ll)0);
printf("%lld\n",ans);
}
return 0;
}