【-30】agc 001F - Wide Swap[拓扑排序]

2017-10-11

Problem Statement

You are given a permutation $P_1…P$ $N$ of the set { $1, 2, …, N.$ }

You can apply the following operation to this permutation, any number of times (possibly zero):

Choose two indices $i$ , $j$ ( $1 \le i \le j \le N$ ), such that $K$ $\le$ $j−i$ and | $P$ $i$ − $P$ $j$ |=1. Then, swap the values of $P$ $i$ and $P$ $j$ .

Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

$2 \le N \le 500,000$
$1\le K \le N−1$
$P$ is a permutation of the set { $1$ , $2$ , …, $N$ }.

Input

The input is given from Standard Input in the following format:

1 2	N K P1 P2 … PN

Output

Print the lexicographically smallest permutation that can be obtained.

Sample Input 1

1 2	4 2 4 2 3 1

Sample Output 1

One possible way to obtain the lexicographically smallest permutation is shown below:

4231
4132
3142
2143
Sample Input 2
1
2
5 1
5 4 3 2 1

Sample Output 2

Sample Input 3

1 2	8 3 4 5 7 8 3 1 2 6

Sample Output 3

题解

读入的 $P_i$ 表示第 $i$ 个位置上的数为 $P_i$

然后设 $Q_i$ 表示 $i$ 这个数在排列中的位置为 $Q_i$

要使交换后 $P$ 数组字典序最小,就是使交换后的 $Q$ 数组字典序最小.(因为使尽量小的数在尽量靠前的地方,就可以使整个排列尽量小)

求出最小字典序后,在 $i$ 前面的数只有比 $i$ 小和比 $i$ 大 $|i-j|>=K$

所以对于 $|i-j|<K$ 的情况, $P_i$ 和 $P_j$ 的相对位置不会变,所以连边跑拓扑排序,求最小的拓扑序

每个点只需要连向左右最小的满足条件的数即可


#include<bits/stdc++.h>
using namespace std;
int read()
{
    char s;
    int k=0,base=1;
    while((s=getchar())!='-'&&s!=EOF&&!(s>='0'&&s<='9'));
    if(s==EOF)exit(0);
    if(s=='-')base=-1,s=getchar();
    while(s>='0'&&s<='9')
    {
        k=k*10+(s-'0');
        s=getchar();
    }
    return k*base;
}
void write(int x)
{
    if(x<0)
    {
        putchar('-');
        write(-x);
    }
    else
    {
        if(x/10)write(x/10);
        putchar(x%10+'0');
    }
}
const int maxn=500010;
const int maxm=1010010;
int n,k,T;
int rd[maxn];
int p[maxn],q[maxn];
int ans[maxn];
int id,to[maxm<<1],ne[maxm<<1],po[maxn];
inline void add(int x,int y)
{
    id++;
    to[id]=y;ne[id]=po[x];
    po[x]=id;
    rd[y]++;
}
set<int> s;
set<int>::iterator it;
priority_queue<int,vector<int>,greater<int> > Q;
int main()
{
    freopen("permutation.in","r",stdin);
    freopen("permutation.out","w",stdout);
    n=read();k=read();
    for (int i=1;i<=n;i++) q[p[i]=read()]=i;
    for (int i=1;i<=n;i++)//前面的
    {
        if (i>k) s.erase(p[i-k]);
        it=s.insert(p[i]).first;//|i-j|<k,Pi<Pj,i和j连边
        if (++it!=s.end()) add(i,q[*it]);
    }
    for (int i=n;i>=1;i--)//后面的
    {
        if (i+k<=n) s.erase(p[i+k]);
        it=s.insert(p[i]).first;
        if (++it!=s.end()) add(i,q[*it]);
    }
    for (int i=1;i<=n;i++)//入度为0的点入队
    {
        if (rd[i]==0) Q.push(i);
    }
    for (int i=1;i<=n;i++)
    {
        T=Q.top();Q.pop();
        ans[T]=i;
        for (int j=po[T];j;j=ne[j])
        {
            rd[to[j]]--;
            if (rd[to[j]]==0) Q.push(to[j]);
        }
    }
    for (int i=1;i<=n;i++)
    {
        printf("%d\n",ans[i]);
    }
    return 0;
}