hdu 4911 Inversion（歸並排序求逆序對數）
Inversion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:
131072/131072 K (Java/Others)
Problem Description
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
i>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
題意：給出n個數，每次可以交換相鄰的兩個數，最多交換k次，求交換後最小的逆序數是多少。
分析：如果逆序數大於0，則存在1 ≤ i < n，使得交換ai和ai+1後逆序數減1。所以最後的答案就是max((inversion-k), 0)。利用歸並排序求出原序列的逆序對數就可以解決問題了。
#include
#include
#define N 100005
__int64 cnt, k;
int a[N],c[N];
//歸並排序的合並操作
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];
}
//歸並排序的遞歸分解和合並
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);
}
}
