HDU4911 Inversion(线段树)

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#算法 #线段树

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本文介绍了一种通过线段树处理序列中的逆序对问题的算法，旨在找到经过有限次数相邻元素交换后逆序对数量的最小值。文章详细阐述了算法原理及其实现过程。

Inversion

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4258 Accepted Submission(s): 1561

Problem 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

Sample Output

1
2

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

题意:给一个序列,你有k次交换相邻两个数的机会,问k此操作之后逆序对的最小值.

题解:用线段树处理出最开始的逆序数,每次交换选择最大的一个往后移,也就是说每次交换都会减少一个逆序对.结果就是max(0,sum-k).

注意要先离散化.

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <queue>
#include <ctype.h>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const long long MAXN=300000+10;
const long long INF=1e9+7;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

long long sum[MAXN<<2];

void pushUp(long long rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void build(long long l,long long r,long long rt)
{
    sum[rt]=0;
    if(l==r)
        return;
    long long m=(l+r)/2;
    build(lson);
    build(rson);
}

void update(long long p,long long l,long long r,long long rt)
{
    if(l==r){
        sum[rt]++;
        return;
    }
    long long m=(l+r)>>1;
    if(p<=m)
        update(p,lson);
    else
        update(p,rson);
    pushUp(rt);
}

long long query(long long L,long long R,long long l,long long r,long long rt)
{
    if(L<=l&&R>=r)
        return sum[rt];
    long long m=(l+r)>>1;
    long long ret=0;
    if(L<=m)
        ret+=query(L,R,lson);
    if(R>m)
        ret+=query(L,R,rson);
    return ret;
}

long long a[MAXN];
long long b[MAXN];
map<long long,long long> m;
int main()
{
    long long n,k;
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        m.clear();
        long long sum=0;
        long long cnt=0;
        for(long long i=0;i<n;i++)
        {
            scanf("%lld",a+i);
            b[i]=a[i];
        }
        sort(b,b+n);
        int tot=1;
        for(int i=0;i<n;i++){
            if(i==0){
                m[b[i]]=tot;
            }
            else if(b[i]==b[i-1]){
                m[b[i]]=tot;
            }else{
                m[b[i]]=++tot;
            }
        }
        build(1,tot,1);
        for(int i=0;i<n;i++){
            a[i]=m[a[i]];
            sum+=query(a[i]+1,tot,1,tot,1);
            update(a[i],1,tot,1);
        }
        if(sum>=k){
            printf("%lld\n",sum-k);
        }else{
            printf("0\n");
        }
    }
}