Andrew Stankevich Contest 28 - F

最新推荐文章于 2024-04-25 09:26:19 发布

SuPhoebe

最新推荐文章于 2024-04-25 09:26:19 发布

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分类专栏：数据结构 & hash 模拟 & 匹配

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本文介绍了一种压缩算法中使用的序列转换方法——Move-to-Front编码，并提供了两种实现方案：一种是直接模拟方法，利用树状数组进行优化；另一种是使用莫队算法结合树状数组来高效解决重复元素的问题。

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Problem F. Move to Front

Input file: mtf.in
Output file: mtf.out
Time limit: 1 second
Memory limit: 256 megabytes

Move-to-Front is a method of transforming sequences of positive integer numbers, that is used in some
compression algorithms, such as Burrows-Wheeler transform.
Initially all positive integer numbers are organized as an ordered list $L$ in their natural order. Consider a sequence $a_1, a_2, . . . , a_n$ of positive integer numbers. It is encoded as a sequence $b_1, b_2, . . . , b_n$ in the following way. Let the part of the sequence from a1 to ai−1 be encoded. The position of ai
in the current list L is considered. It is assigned to bi
, and $a_i$ is moved to the beginning of the list $L$ .
For example, the sequence $3, 3, 3, 2, 2, 2, 2, 2, 3, 1, 3, 3, 2$ is encoded as $3, 1, 1, 3, 1, 1, 1, 1, 2, 3, 2, 1, 3$ .
You are given a sequence $a_1, a_2, . . . , a_n$ , you must encode it using Move-to-Front method, and output the resulting sequence $b_1, b_2, . . . , b_n$ .

Input

The first line of the input file contains integer number $n (1 ≤ n ≤ 100 000)$ . The second line contains n
integer numbers $a_i$ , ranging from $1 ～ 10^9$ .

Output

Output $n$ integer numbers — the sequence $b_1, b_2, . . . , b_n$ .

Example

mtf.in
13
3 3 3 2 2 2 2 2 3 1 3 3 2
mtf.out
3 1 1 3 1 1 1 1 2 3 2 1 3

直接模拟
将数字离散化，配合扩大两倍的树状数组
从后面开始删，直接插入至前面位置，可以证明这样花费的空间至多是 $2n$ ，所以对于离散化之后最大的空间不超过20w。

//      whn6325689
//      Mr.Phoebe
//      http://blog.youkuaiyun.com/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;

#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))

#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))

#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
#define root 1,1,n

template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------

const int MAXN=100010;

int n,temp;
int cnt;
int a[MAXN],b[MAXN],pos[MAXN];
int lis[MAXN],sum=0;

int t[MAXN*4];
void add(int i,int v)
{
    for(;i<MAXN*4;i+=lowbit(i))
        t[i]+=v;
}
int getsum(int i)
{
    int sum=0;
    for(;i;i-=lowbit(i))
        sum+=t[i];
    return sum;
}

int main()
{
    freopen("mtf.in","r",stdin);
freopen("mtf.out","w",stdout);
    read(n);
    for(int i=1;i<=n;lis[i]=a[i],i++)
        read(a[i]);
    sort(lis+1,lis+n+1);
    cnt=unique(lis+1,lis+n+1)-lis;
    for(int i=1;i<=n;i++)
        a[i]=lower_bound(lis+1,lis+cnt,a[i])-lis;
    lis[0]=0;
    int now=400000;
    for(int i=cnt;i>=1;i--)
    {
        pos[i]=now--;
        add(pos[i],1);
        if(lis[i]-lis[i-1]>=2)
        {
            add(now,lis[i]-lis[i-1]-1);
            now--;
        }
    }

    for(int i=1;i<=n;i++)
    {
        printf("%d ",getsum(pos[a[i]]));
        add(pos[a[i]],-1);
        pos[a[i]]=now--;
        add(pos[a[i]],1);
    }

    return 0;
}

莫队算法
对于第一次出现的数字，我们用树状数组记录有多少种大于此时的数的数字出现。
对于第二次之后出现的数字，通过莫队算法计算上一次出现了这个数的位置到现在这个位置中间出现过多少种数，因为出现了就变成第一位了，中间出现过多少种数就是 $b_i$ 的值。

//      whn6325689
//      Mr.Phoebe
//      http://blog.youkuaiyun.com/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;

#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))

#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))

#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
#define root 1,1,n

template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------

const int MAXN=100010;

int n,temp;
int cnt;
int a[MAXN],b[MAXN],c[MAXN];
int lis[MAXN],sum=0;
int vis[MAXN],fff[MAXN];

int t[MAXN];
void add(int i)
{
    for(;i<n+10;i+=lowbit(i))
        t[i]+=1;
}
int getsum(int i)
{
    int sum=0;
    for(;i;i-=lowbit(i))
        sum+=t[i];
    return sum;
}

int last[MAXN];
int ans[MAXN];
struct yyy
{
    int x,y;
};
int Size;
vector<yyy> g;

int cmp(yyy x,yyy y)
{
    if (x.x / Size == y.x / Size)
        return x.y < y.y;
    else return x.x < y.x;
}

int num[MAXN];

void ins(int x)
{
    num[x]++;
    if (num[x] == 1) sum++;
}
void era(int x)
{
    num[x]--;
    if (num[x] == 0) sum--;
}

int main()
{
    freopen("mtf.in","r",stdin);
    freopen("mtf.out","w",stdout);
    read(n);
    for(int i=1;i<=n;lis[i]=a[i],i++)
        read(a[i]);
    sort(lis+1,lis+n+1);
    cnt=unique(lis+1,lis+n+1)-lis-1;
    for(int i=1;i<=n;i++)
    {
        temp=lower_bound(lis+1,lis+cnt+1,a[i])-lis;
        if(vis[temp]){a[i]=temp;continue;}
        fff[temp]=a[i]+sum-getsum(temp);
        a[i]=temp;add(temp);vis[temp]=1;sum++;
    }
    /*for(int i=1;i<=cnt;i++)
        cout<<fff[i]<<" ";*/
    yyy tmp;
    for (int i = 1 ; i <= n ;i++)
    {
        if (last[a[i]] == 0)
        {
            ans[i] = fff[a[i]];
        }
        else
        {
            tmp.x = last[a[i]];
            tmp.y = i;
            g.push_back(tmp);
        }
        last[a[i]] = i;
    }
    Size = sqrt(g.size());
    sum = 0;
    sort(g.begin(),g.end(),cmp);

    int l = 1,r = 1;

    num[a[1]] = 1;
    sum = 1;

    for (int i = 0 ; i < g.size() ; i++)
    {
        while (r < g[i].y)
        {
            r++;
            ins(a[r]);
        }

        while (r > g[i].y)
        {
            era(a[r]);
            r--;
        }
        while (l < g[i].x)
        {
            era(a[l]);
            l++;
        }
        while (l > g[i].x)
        {
            l--;
            ins(a[l]);
        }
        //cout<<l<<' '<<r<<' '<<sum<<endl;
        ans[r] = sum ;

    }
    for (int i= 1 ; i <= n ;i++)
        printf("%d ",ans[i]);

    return 0;
}