HDU - 4911 Inversion(归并排序求逆序数)

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题意:给一个长度为n的序列,问经过最多k次两两交换(仅限相邻的数之间),最少还有多少逆序对.

分析:归并求出总的逆序对数,然后ans=max(求出的总逆序对-k,0).

哇&#$%,Anything is possible.

参考代码:

#include
#include
#include
#include
#include

using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
int n;
LL k;
int a[maxn];
int tmp[maxn];
LL ans;//emmmmmm数据有点大 用LL才能过

void Merge( int l, int mid, int r)
{
    int i = l;
    int j = mid+1;
    int k = l;
    while( i <= mid && j <= r)
    {
        if( a[i] > a[j])
        {
            tmp[k++] = a[j++];
            ans += mid-i+1;
        }
        else
            tmp[k++] = a[i++];
    }

    while( i <= mid)
        tmp[k++] = a[i++];
    while( j <= r)
        tmp[k++] = a[j++];
    for( int i = l; i <= r; i++)
        a[i] = tmp[i];
}

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

int main()
{
    while( ~scanf("%d%lld",&n,&k))
    {
        for( int i = 0; i < n; i++)
            scanf("%d",&a[i]);
        ans = 0;
        MergeSort(0,n-1);
        printf("%lld\n",max(0LL,ans-k));
    }

    return 0;
}