hdu 4911 Inversion（找到的倒数）

主题链接：http://acm.hdu.edu.cn/showproblem.php?pid=4911

Inversion

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

Total Submission(s): 528 Accepted Submission(s): 228

[align=left]Problem Description[/align]
bobo has a sequence a1,a2,…,an. He is allowed to swap twoadjacent 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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each tests:

A single integer denotes the minimum number of inversions.

[align=left]Sample Input[/align]

3 1
2 2 1
3 0
2 2 1

[align=left]Sample Output[/align]

1
2

[align=left]Author[/align]
Xiaoxu Guo (ftiasch)

[align=left]Source[/align]
2014 Multi-University Training Contest 5

解法：求的所给序列的逆序数。然后减掉k，假设小于零就去零！

代码例如以下：（归并排序法）

#include<stdio.h>

int is1[112345],is2[112345];// is1为原数组，is2为暂时数组。n为个人定义的长度

__int64 merge(int low,int mid,int high)
{
int i=low,j=mid+1,k=low;
__int64 count=0;
while(i<=mid&&j<=high)
if(is1[i]<=is1[j])// 此处为稳定排序的关键，不能用小于
is2[k++]=is1[i++];
else
{
is2[k++]=is1[j++];
count+=j-k;// 每当后段的数组元素提前时。记录提前的距离
}
while(i<=mid)
is2[k++]=is1[i++];
while(j<=high)
is2[k++]=is1[j++];
for(i=low;i<=high;i++)// 写回原数组
is1[i]=is2[i];
return count;
}
__int64 mergeSort(int a,int b)// 下标，比如数组int is[5]，所有排序的调用为mergeSort(0,4)
{
if(a<b)
{
int mid=(a+b)/2;
__int64 count=0;
count+=mergeSort(a,mid);
count+=mergeSort(mid+1,b);
count+=merge(a,mid,b);
return count;
}
return 0;
}
int main()
{
int n, x;
__int64 k;
__int64 sum;
while(scanf("%d%I64d",&n,&k)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%d",&x);
is1[i] = x;
}
__int64 ans=mergeSort(0,n-1);
sum=0;
printf("%I64d\n",ans-k>0?ans-k:0);
}
return 0;
}