hdu 4911求逆序数

如果逆序数大于0，存在1<=i<n，使得交换两个数数后逆序数减少1.

所以答案为max{inversion-k,0}.

经典的分治法复杂度为n*log(n).

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>a _j.

Input

The 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,a₂,…,a _n (0≤a _i≤10 ⁹).

Output

For each tests: 

A single integer denotes the minimum number of inversions.

Sample Input

3 12 2 13 02 2 1

Sample Output

1
2
#define N 100005long long cnt, k;int a[N],c[N];//归并排序的合并操作void merge(int a[], int first, int mid, int last, int c[]){    int i = first, j = mid + 1;    int m = mid, n = last;    int k = 0;    while(i <= m || j <= n)    {        if(j > n || (i <= m && a[i] <= a[j]))            c[k++] = a[i++];        else        {            c[k++] = a[j++];            cnt += (m - i + 1);        }    }    for(i = 0; i < k; i++)        a[first + i] = c[i];}//归并排序的递归分解和合并void merge_sort(int a[], int first, int last, int c[]){    if(first < last)    {        int mid = (first + last) / 2;        merge_sort(a, first, mid, c);        merge_sort(a, mid+1, last, c);        merge(a, first, mid, last, c);    }}int main(){    int n;    while(~scanf("%d%lld",&n,&k))    {        memset(c, 0, sizeof(c));        cnt = 0;        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        merge_sort(a, 0, n-1, c);        if(k >= cnt) cnt = 0;        else cnt -= k;        printf("%lld\n",cnt);    }    return 0;}

树状数组做法
#include <cstdio>  #include <cstring>  #include <algorithm>    using namespace std;    const int MAXN = 100000 + 10;    int n, k;  int a[MAXN], b[MAXN];  long long C[MAXN];    void Init()  {      memset(C, 0, sizeof(C));  }    void Read()  {      for (int i = 0; i < n; ++i)      {          scanf("%d", a + i);      }  }    int Lowbit(int x)  {      return x & (-x);  }    long long Sum(int x)  {      long long nRet = 0;        while (x > 0)      {          nRet += C[x];          x -= Lowbit(x);      }        return nRet;  }    void Add(int x)  {      while (x <= n)      {          C[x]++;          x += Lowbit(x);      }  }    void Solve()  {      int nCnt = 0;      int nId = 0;      long long nReverse = 0;        memcpy(b, a, sizeof(a));      sort(b, b + n);      nCnt = unique(b, b + n) - b;        for (int i = n - 1; i >= 0; --i)      {          nId = lower_bound(b, b + nCnt, a[i]) - b + 1;          nReverse += Sum(nId - 1);          Add(nId);      }        if (nReverse > k)      {          nReverse -= k;      }      else      {          nReverse = 0;      }        printf("%I64d\n", nReverse);  }    int main()  {      while (scanf("%d%d", &n, &k) == 2)      {          Init();          Read();          Solve();      }      return 0;  }