[HDOJ4911]Inversion

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): 2744    Accepted Submission(s): 1015

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 1 2 2 1 3 0 2 2 1

 

Sample Output

1 2

　　求交换k次以后所获得的数列的最小的逆序数。

　　如果逆序数大于0，说明数列中有两个数可以交换，使得逆序数-1。所以所求交换k次所得数列最小逆序数的结果就是排序结束后交换次数-k，这个结果大于等于0。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

typedef long long LL;
const int maxn = 100010;
int num[maxn];
int Right[maxn], Left[maxn];
LL ans;

void merge(int* num, int p, int q, int r) {
    int n1, n2, i, j, k;
    n1 = q - p + 1;
    n2 = r - q;
    for(i = 0; i < n1; i++) {
        Left[i] = num[p+i];
    }
    for(i = 0; i < n2; i++) {
        Right[i] = num[q+i+1];
    }
    Left[n1] = Right[n2] = 0x7FFFFFFF;
    i = 0;
    j = 0;
    for(k = p; k <= r; k++) {
        if(Left[i] <= Right[j]) {
            num[k] = Left[i];
            i++;
        }
        else {
            num[k] = Right[j];
            j++;
            ans += (n1 - i);
        }
    }
}

void mergesort(int* num, int p, int r) {
    int q;
    if(p < r) {
        q = (p + r) / 2;
        mergesort(num, p, q);
        mergesort(num, q+1, r);
        merge(num, p, q, r);
    }
}

int main() {
    int n, k;
    while(~scanf("%d %d", &n, &k)) {
        ans = 0;
        for(int i = 0; i < n; i++) {
            scanf("%d", &num[i]);
        }
        mergesort(num, 0, n-1);
        cout << ((ans-k) > LL(0) ? ans-k : 0) << endl;
    }
}