HDU 4911 Inversion(歸併排序求逆序數)

畫船聽雨發表於2014-08-06
排序

歸併排序求逆序數，然後ans-k與0取一個最大值就可以了。

也可以用樹狀陣列做，比賽的時候可能姿勢不對，樹狀陣列wa了、、

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 578    Accepted Submission(s): 249


Problem Description
bobo has a sequence a1,a2,…,an. 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 ai>aj.
 

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,…,an (0≤ai≤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





 


#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;


const int maxn = 10010000;

int num[maxn];
LL ans;
int tmp[maxn];

void Sell(int l, int mid, int r)
{
    int i = l;
    int j = mid+1;
    int k = 0;
    while(i <= mid && j <= r)
    {
        if(num[i] > num[j])
        {
            tmp[k++] = num[j++];
            ans += mid-i+1;
        }
        else tmp[k++] = num[i++];
    }
    while(i <= mid)
        tmp[k++] = num[i++];
    while(j <= r)
        tmp[k++] = num[j++];
    for(int i = 0; i < k; i++)
        num[i+l] = tmp[i];
}

void Sort(int l, int r)
{
    if(l < r)
    {
        int mid = (l+r)/2;
        Sort(l, mid);
        Sort(mid+1, r);
        Sell(l, mid, r);
    }
}
int main()
{
    int n, k;
    while(cin >>n>>k)
    {
        ans = 0;
        for(int  i = 0;i < n; i++) scanf("%d",&num[i]);
        Sort(0, n-1);
        cout<<max(0LL, ans-k)<<endl;
    }
    return 0;
}




            




相關文章