hdu 4911 Inversion（求逆序数）

题目链接：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

Problem Description

bobo has a sequence a₁,a₂,…,a_n. 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 a_i>a_j.

 

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,…,a_n (0≤a_i≤10⁹).

 

Output

For each tests:

A single integer denotes the minimum number of inversions.

 

Sample Input

3 12 2 13 02 2 1

 

Sample Output

12

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5 

解法：求的所给序列的逆序数，然后减掉k，如果小于零就去零！

代码如下：（归并排序法）

#include<stdio.h>int is1[112345],is2[112345];// is1为原数组，is2为临时数组，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，例如数组int is[5]，全部排序的调用为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;}