HDU 4911 Inversion （逆序数归并排序）

Inversion

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121349#problem/A

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<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,a 2,…,a n (0≤a i≤10 9).

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

##题意：对有n个元素的数组进行不超过k次操作，每次操作允许交换相邻元素; 求操作后能得到的整个数组的最小逆序数.
##题解：容易得出每次交换相邻元素最多只能使得逆序数降低1. 而对于一个逆序数不为0的数组，一定存在相邻的两个数满足(i < j, ai < aj)(可反证). 所以每次操作都可以使得逆序数减少1. 最终答案即为max(0, 逆序数). 求逆序数可以用归并排序、树状数组和线段树(数的规模较小时).
##代码： ``` cpp #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 101000 #define mod 100000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

LL num[maxn];
LL _count=0;

void merge_array(int left,int mid,int right)
{
int temp[right-left+1];int k=0;
int i=left;int j=mid+1;
while(i<=mid&&j<=right)
{
if(num[i]<=num[j])
temp[k++]=num[i++];
else
{temp[k++]=num[j++];_count+=mid-i+1;}
}
while(i<=mid)
temp[k++]=num[i++];
while(j<=right)
temp[k++]=num[j++];

for(i=left;i<=right;i++)
    num[i]=temp[i-left];

}

void merge_sort(int left,int right)
{
if(left<right)
{
int mid=left+(right-left)/2;
merge_sort(left,mid);
merge_sort(mid+1,right);
merge_array(left,mid,right);
}
}

int main(void)
{
//IN;

int n;
LL k;
while(scanf("%d %I64d",&n,&k) != EOF)
{
    for(int i=0; i<n; i++)
        scanf("%I64d",&num[i]);
    _count=0;
    merge_sort(0, n-1);

    printf("%I64d
", max(_count-k,0LL));
}
return 0;

}

HDU 4911 Inversion （逆序数 归并排序）