﻿HDU 4911 Inversion-Ocrosoft

# HDU 4911 Inversion

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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```

Solution

```题意：交换两个相邻的数不超过k次，求交换后的最小逆序对数。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
const int INF = INT_MAX / 2;
const int MAXN = 100000 + 10;
using namespace std;
//逆序对模板
int a[100005];
int le[100005], ri[100005];
ll cnt;
void merge(int* a, int p, int q, int r)
{
int i, j, k, n1, n2;
n1 = q - p + 1;
n2 = r - q;
for (i = 0; i < n1; i++)
{
le[i] = a[p + i];
}
for (i = 0; i < n2; i++)
{
ri[i] = a[q + i + 1];
}
le[n1] = ri[n2] = 0x7fffffff;

i = j = 0;
for (k = p; k <= r; k++)
{
if (le[i] <= ri[j])
{
a[k] = le[i];
i++;
}
else
{
a[k] = ri[j];
j++;
cnt += n1 - i;
}
}
return;
}
void mergesort(int* a, int p, int r)
{
int q;
if (p < r)
{
q = (p + r) / 2;
mergesort(a, p, q);
mergesort(a, q + 1, r);
merge(a, p, q, r);
}
return;
}
int main()
{
int n, k;
while (cin >> n >> k)
{
cnt = 0;
for (int i = 0; i < n; i++)scanf("%d", &a[i]);
mergesort(a, 0, n - 1);
//如果逆序数大于0，那么必定存在1 <= i<n使得i和i + 1交换后逆序数减1
printf("%I64d\n", max(cnt - k, (ll)0));
}
return 0;
}

```