4thweek contest.A 归并排序求逆序数

题目：

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.

 

分析：

应用二分法排序并记录逆序数k，由于题目一开始输入一个ka,所以交换后的数组的逆序数=k-ka.但需要注意：k和ka的大小问题。具体在代码中解释

 

Input

The input consists of several tests. For each tests: 
The first line contains 2 integers n,ka, (1≤n≤10 ⁵,0≤ka≤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

 

 代码及重要步骤分析：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=100005;
long long a[maxn],b[maxn];
long long k;

int merge_sort(long long x,long long y) //
{
    if(x<y)
    {
        long long m=(x+y)/2; // 划分
        merge_sort(x,m);      //递归求解
        merge_sort(m+1,y);    //递归求解
        int p=0;
        int i=x;
        int j=m+1;
        while(i<=m && j<=y)
        {
            if(a[i]>a[j])
            {
                b[p++]=a[j++];   //从左半数组复制到临时空间
                k+=m+1-i;        //记录逆序数k
            }
            else
                b[p++]=a[i++];   //从右半数组复制到临时空间

        }
        while(i<=m)
            b[p++]=a[i++];
        while(j<=y)
            b[p++]=a[j++];
        for(i=0;i<p;++i)
            a[x+i]=b[i];      //从辅助空间复制回数组a

    }
    return 0;
}

int main()
{
    long long n,ka;
    int i;
    while(scanf("%lld%lld",&n,&ka) !=EOF )  //不要加上&& n

    {
        k=0;
        for(i=0;i<n;i++)
            scanf("%lld",&a[i]);
        merge_sort(0,n-1);
        if(k<=ka)     //如果输入的数组本身就有一定顺序，可能逆序数就会小于需要交换的次数
          printf("0
");
        else
            printf("ll%d
",k-ka);
    }

    return 0;
}