HDU 4911

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5250    Accepted Submission(s): 1845

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 12 2 13 02 2 1

 

Sample Output

12

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5 

 

题意：

给你一个排序，你可以交换k次相邻的数，输出最好方案的最少的逆序数对。

POINT：

容易可知，每次交换都可以减少一个逆序数对。

所以只要求出原始的逆序数对，减去k，再与0比较一下就好了。

求逆序数对

1:因为数比较大，但是个数才1e5，用离散化后树状数组。

2:归并排序。

【训练赛的时候，写树状数组写错TLE，写线段树数组开小了，错了8发，真要反省一下。那么简单的题】

归并：

#include <string>#include <string.h>#include <iostream>#include <queue>#include <math.h>#include <stdio.h>using namespace std;#define LL long longconst LL maxn = 1e5+55;LL a[maxn];LL ans=0;void merge(LL l,LL mid,LL r){LL now=0;LL ll=l;LL rr=mid+1;LL cnt[maxn];while(ll<=mid&&rr<=r){if(a[ll]<=a[rr]){cnt[++now]=a[ll++];}else{cnt[++now]=a[rr++];ans+=mid-ll+1;}}while(ll<=mid){cnt[++now]=a[ll++];}while(rr<=r){cnt[++now]=a[rr++];}for(LL i=1;i<=now;i++){a[l+i-1]=cnt[i];}}void bf(LL l,LL r){if(l==r) return;LL mid=(l+r)>>1;bf(l,mid);bf(mid+1,r);merge(l,mid,r);}int main(){LL n,k;while(~scanf("%lld %lld",&n,&k)){ans=0;for(LL i=1;i<=n;i++) scanf("%lld",&a[i]);bf(1,n);printf("%lld\n",max(ans-k,1LL*0));}    return 0;}

树状数组：

#include<iostream>#include<iomanip>#include <cstdio>#include <string.h>#include <algorithm>#include <stack>#include <cmath>using namespace std;typedef long long LL;const LL maxn = 2*1e5+33;LL a[maxn];LL now[maxn];LL sum[maxn];LL n,k;LL cnt=0;LL lowbit(LL x){    return x&-x;}void add(LL x,LL p){    for(LL i=x;i<=cnt;i+=lowbit(i)){        sum[i]++;    }}LL query(LL x){    LL ans=0;    for(LL i=x;i>=1;i-=lowbit(i)){        ans+=sum[i];    }    return ans;}int main(){    while(~scanf("%lld %lld",&n,&k)){        cnt=0;        for(LL i=1;i<=n;i++){            sum[i]=0;            scanf("%lld",&a[i]);            now[i]=a[i];        }        sort(now+1,now+1+n);        now[0]=-111;        for(LL i=1;i<=n;i++){            if(now[i]==now[i-1]) continue;            now[++cnt]=now[i];        }        LL ans=0;        for(LL i=1;i<=n;i++){            a[i]=(LL)(lower_bound(now+1,now+1+cnt,a[i])-now);            add(a[i],1);            ans+=query(cnt)-query(a[i]);        }        ans=max(1LL*0,ans-k);        printf("%lld\n",ans);    }    return 0;}