ACM 逆序对 hdu 4911 Inversion 离散化 树状数组模板

ACM 逆序对

简单可证明，每一次移动都可以减少一个逆序对，因此

设逆序对个数为sum,这答案为sum-k，如sum-k<0则直接输出0即可

先排序离散化，然后逆序对使用树状数组求。

Inversion

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 244 Accepted Submission(s): 102



Problem Description
bobo has a sequence a1,a2,…,an. He is allowed to swap twoadjacent 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 ai>aj.


Input
The input consists of several tests. For each tests:

The first line contains 2 integers n,k (1≤n≤105,0≤k≤109). The second line contains n integers a1,a2,…,an (0≤ai≤109).


Output
For 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

Author
Xiaoxu Guo (ftiasch)

Accepted

1001

250MS

2332K

G++

#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
#define out(x) cout<<#x<<": "<<x<<endl
const double eps(1e-8);
const int maxn=101000;
const long long maxint=-1u>>1;
const long long maxlong=maxint*maxint;
typedef long long lint;
struct wjj
{
    int x,y;
};
int n,k,b[maxn],big;
long long f[maxn];
wjj a[maxn];
long long sum;

void init()
{
    for (int i=1; i<=n; i++) 
    {
        scanf("%d",&a[i].x);
        a[i].y=i;
    }
}
bool cmp(wjj a,wjj b)
{
    return (a.x<b.x || (a.x==b.x && a.y<b.y));
}

int lowb(int x){
    return x&(-x);
}

long long find(int x){
    long long tmp = 0;
    int y = x;
    while (y>0){
        tmp += f[y];
        y -= lowb(y);
    }
    return tmp;
}

void insert(int x){
    int y = x;
    while (y<=n){
        f[y] ++;
        y += lowb(y);
    }
}

void work()
{
    big=0;
    sort(a+1,a+1+n,cmp);
    a[0].x=-1;
    for (int i=1; i<=n; i++)
    {
        if (a[i].x!=a[i-1].x) big++;
        b[a[i].y]=big;
    }
    memset(f,0,sizeof(f));
    long long ans = 0;
    for (int i = 1; i <= n; i ++){
        ans +=  (find(n)-find(b[i]));
        insert(b[i]);
    }
    if (ans<=k) cout<<0<<endl;
    else cout<<ans-k<<endl;
}

int main()
{
    while(cin>>n>>k)
    {
        init();
        work();
    }
    return 0;
}