hdu 4911 Inversion

Inversion

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

Total Submission(s): 2222 Accepted Submission(s): 859



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≤i<j≤n and ai>aj.

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

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

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题意:输入n,k n为序列的长度,k为交换的次数,每次只能是相邻元素之间的交换,求该序列经过k次交换得到最小的逆序数。

若1个序列的逆序数大于0,总存在0<=i<n使得 swap(ai,ai+1)之后 逆序数减1；

因此用归并排序求出序列的逆序数cnt与k进行比较即可得到答案

还要注意 用int存储逆序数会数据溢出

AC代码：

#include <stdio.h> #include "string.h" #include "math.h" long long int cnt; void Merge(int ss[],int ex,int ey,int sx,int sy) { int temp[100002],i=ex,j=sx,pos=ex; for(;i<=ey&&j<=sy;) { if(ss[i]<=ss[j]) temp[pos++]=ss[i++]; else { temp[pos++]=ss[j++]; cnt+=(ey-i+1); } } while(i!=ey+1) temp[pos++]=ss[i++]; while(j!=sy+1) temp[pos++]=ss[j++]; for(i=ex;i<=sy;i++) { ss[i]=temp[i]; } } void Mergesort(int ss[],int left,int right) { int middle; if(left<right) { middle=(left+right)/2; Mergesort(ss,left,middle); Mergesort(ss, middle+1, right); Merge(ss,left,middle,middle+1,right); } } int main() { int n,k; while(scanf("%d%d",&n,&k)!=EOF) { int ss[100002]; cnt=0; for(int i=0;i<n;i++) scanf("%d",&ss[i]); Mergesort(ss, 0, n-1); if(k>cnt) printf("0\n"); else printf("%lld\n",cnt-k); } }