hdu 4911 Inversion

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≤iaj.

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

思路：归并排序求逆序数。
把原数组化成每个长度为1的小块，然后再合并，再合并中，每次当后面的数小于前面的数时，逆序数（cnt+=mid-i+1）,因为每个数组自生有序，所以是从中间开始寻找逆序数的个数。

 #include
using namespace std;

typedef long long ll;
const int maxn = 100007;
ll a[maxn], b[maxn];//临时存放变量
ll cnt = 0;
void Merge(ll l,ll mid,ll r)
{
	int i = l, j = mid + 1, t = 0;//t是b数组的下标
	while(i<=mid&&j<=r)
	{
		if(a[i]>a[j])
			b[t++] = a[j++], cnt += mid - i + 1;//计算逆序数
		else
			b[t++] = a[i++];
	}
	while(i<=mid)	b[t++] = a[i++];
	while(j<=r)	b[t++] = a[j++];
	for (int i = 0;i<t;i++)
		a[l + i] = b[i];//把b[]中的值复制给a[]
}

void Mergesort(ll l,ll r)
{
	if(l<r)
	{
		ll mid = (l + r) >> 1;
		Mergesort(l, mid);
		Mergesort(mid+1, r);
		Merge(l,mid,r);
	}
}

int main()
{
	freopen("11.txt", "r", stdin);
	ll n, k;
	while(scanf("%lld%lld",&n,&k)!=EOF)
	{
		cnt = 0;
		for (int i = 0;i<n;i++)
			scanf("%lld", &a[i]);
		Mergesort(0, n-1);
		if(cnt<=k)
			printf("0\n");
		else
			printf("%lld\n", cnt - k);
	}

	return 0;
}

归并排序代码：

#include
using namespace std;
//归并排序
typedef long long ll;
const int maxn = 10000;
ll a[maxn], b[maxn];

void Merge(ll l,ll mid,ll r)
{
	ll i = l, j = mid + 1, t = 0;
	while(i<=mid&&j<=r)
	{
		if(a[i]>a[j])
			b[t++] = a[j++];
		else
			b[t++] = a[i++];
		//cnt += mid - i + 1;
	}
	while(i<=mid)	b[t++] = a[i++];
	while(j<=r)	b[t++] = a[j++];
	for (int i = 0;i<t;i++)
		a[l + i] = b[i];//把排序号的b[]复制给a[]
}


void Mergesort(ll l,ll r)
{
	if(l<r)
	{
		ll mid = (l + r) >> 1;
		Mergesort(l, mid);
		Mergesort(mid + 1, r);
		Merge(l, mid, r);
	}
}

int main()
{
	a[0] = 4;
	a[1] = 1;
	a[2] = 3;
	a[3] = 2;
	a[4] = 4;
	a[5] = 8;
	a[6] = 15;
	Mergesort(0, 6);
	for (int i = 0;i<=6;i++)
		cout << a[i] << " ";

	return 0;
}