HDOJ 4911-Inversion【逆序对模板】

Inversion

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

Total Submission(s): 3201 Accepted Submission(s): 1173



Problem Description

bobo has a sequence a1,a2,…,an. 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 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)

解题思路：

题意：给出n个数，每次可以交换相邻的两个数，最多交换k次，求交换后最小的逆序数是多少。

分析：如果逆序数大于0，则存在1 ≤ i < n，使得交换ai和ai+1后逆序数减1。所以最后的答案就是max((inversion-k), 0)。利用归并排序求出原序列的逆序对数就可以解决问题了。

#include<stdio.h>
#include<string.h>
#define N 100005
__int64 cnt, k;
int a
,c
;
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];
}
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%I64d",&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("%I64d\n",cnt);
}
}