HDU 4911

Description

bobo has a sequence a ₁,a ₂,…,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≤10 ⁵,0≤k≤10 ⁹). The second line contains n integers a ₁,a ₂,…,a _n (0≤a _i≤10 ⁹).

 

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

 

 

题意：给一个序列和交换次数k，求交换后序列最小的的逆序数....

 

 

解题思路：先求出原始序列的逆序数，然后再减去交换次数，就是答案...值得注意的是，如果交换次数减原始逆序数小于0则，答案为0...

 

 

求逆序数，这里用到了归并排序..

 

 

 

 

代码如下：0

 

 

 

#include <stdio.h>
#include <string.h>
int n,k,A[100000],T[100000];
long long ans;
void guibing(int* A,int x,int y,int* T)
{
    if(y-x>1){
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        guibing(A,x,m,T);
        guibing(A,m,y,T);
        while(p<m||q<y){
            if(q>=y||(p<m&&A[p]<=A[q])) T[i++]=A[p++];
            else {
                T[i++]=A[q++];
                ans+=m-p;
            }
        }
        for(i=x;i<y;i++) A[i]=T[i];
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)==2&&n){
        ans=0;
        for(int i=0;i<n;i++){
            scanf("%d",&A[i]);
        }
        memset(T,0,sizeof(T));
        guibing(A,0,n,T);
        if(ans-k<0)
            ans=0;
        else
            ans=ans-k;
        printf("%I64d\n",ans);
    }
    return 0;
}