Inversion HDU - 4911 (逆序对)

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).OutputFor 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：给你一个序列，任意交换两个相邻元素，不超过k次，在交换之后，问最少的逆序对有多少个？

思路：假设逆序数大于0，存在0 <= i < n使得交换Ａi，Ａi+1后逆序数降低1，所求答案就为max(ans - k, 0);

利用归并排序计算逆序对数。

 

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
#include <string>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
//#include <unordered_map>
#define Fbo friend bool operator < (node a, node b)
#define mem(a, b) memset(a, b, sizeof(a))
#define FOR(a, b, c) for (int a = b; a <= c; a++)
#define RFOR(a, b, c) for (int a = b; a >= c; a--)
#define off ios::sync_with_stdio(0)
#define sc(a) scanf("%d",&a) 
bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; }

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int Maxn = 2e5 + 9;
const double pi = acos(-1.0);
const double eps = 1e-8;

ll a[Maxn], b[Maxn];
ll ans;//逆序对数量

void msort(ll l, ll r) {
    if (l == r)return;
    int mid = (l + r) >> 1;
    ll i = l, j = mid + 1, k = l;  //注意这里i = l k = l （是L而不是1）
    msort(l, mid), msort(mid + 1, r);
    while (i <= mid && j <= r) {
        if (a[i] <= a[j])
            b[k++] = a[i++];
        else {
            b[k++] = a[j++];
            ans += (mid - i + 1); //记录逆序对数量
        }
    }
    //一个子序列中的数都处理完了，另一个还没有，把剩下的直接复制过来
    while (i <= mid)b[k++] = a[i++];
    while (j <= r)b[k++] = a[j++];
    for (i = l; i <= r; i++) //把排好序的b[]复制回a[]
        a[i] = b[i];
}

int main() {
    off;
    int n, k;
    while (cin >> n >> k) {
        ans = 0;
        FOR(i, 1, n) sc(a[i]);
        msort(1, n);
        if (ans <= k)cout << 0 << endl;
        else cout << ans - k << endl;
    }
}