hdu 4911 Inversion--2014 Multi-University Training Contest 5

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): 304    Accepted Submission(s): 133

Problem Description

bobo has a sequence a₁,a₂,…,a_n. 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 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

 

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Statistic | Submit | Discuss | Note

这一道是签到题。因为每次操作都会使逆序数减少。只需要用归并排序算出逆序数，然后判断它与k的大小即可。

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>#include<cstdlib>#include<map>#include<set>#include<vector>#include<string>#include<stack>#include<queue>#include<bitset>using namespace std;#define CLR(A)  memset(A,0,sizeof(A))typedef long long ll;const int MAX=500000;int L[MAX],R[MAX],a[MAX];void merge(int *a,int p,int mid ,int q,ll &cnt){    int i,j;    for(i=p;i<=mid;i++) L[i]=a[i];    for(j=mid+1;j<=q;j++) R[j]=a[j];    i=p;j=mid+1;    int w=p;    while(i<=mid&&j<=q){        if(L[i]<=R[j]) a[w++]=L[i++];        else{            cnt+=mid-i+1;            a[w++]=R[j++];        }    }    while(i<=mid) a[w++]=L[i++];    while(j<=q) a[w++]=R[j++];}void mergeSort(int *a,int p,int q,ll &cnt){    if(p<q){        int mid=(p+q)>>1;        mergeSort(a,p,mid,cnt);        mergeSort(a,mid+1,q,cnt);        merge(a,p,mid,q,cnt);    }}int main(){    int n,k;    while(~scanf("%d%d",&n,&k)){        ll cnt=0;        for(int i=0;i<n;i++) scanf("%d",&a[i]);        mergeSort(a,0,n-1,cnt);        cnt=cnt<=k?0:cnt-k;        printf("%I64d\n",cnt);    }    return 0;}