动态规划——Bubble Sort Graph

最新推荐文章于 2025-07-22 00:50:53 发布

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本文介绍了一种基于Bubble Sort算法构建图的方法，并通过动态规划结合二分查找求解该图的最大独立集问题。文章提供了完整的C++实现代码。

Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a₁, a₂, ..., a_n in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).

procedure bubbleSortGraph()
    build a graph G with n vertices and 0 edges
    repeat
        swapped = false
        for i = 1 to n - 1 inclusive do:
            if a[i] > a[i + 1] then
                add an undirected edge in G between a[i] and a[i + 1]
                swap( a[i], a[i + 1] )
                swapped = true
            end if
        end for
    until not swapped 
    /* repeat the algorithm as long as swapped value is true. */ 
end procedure

For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 10⁵). The next line contains n distinct integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n).

Output

Output a single integer — the answer to the problem.

Example

Input

3
3 1 2

Output

Note

Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2].

题意：看了之后才知道是求最长非递减子序列。。。

思路：看数据范围知道是用dp+二分

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[100010];
int dp[100010];
int n;
int wtf(int l, int x)   //二分搜索求dp数组中<=x的位置
{
    int left=1,right=l,mid=(left+right)>>1;
    while(left<=right)
    {
        if(dp[mid]>x)
            right=mid-1;
        else
            if(dp[mid]<x)
                left=mid+1;
            else
                return mid;
        mid=(left+right)>>1;
    }
    return left;
}
int DP()
{
    int l=1;
    dp[1]=a[0];
    for(int i=1; i<n; i++)
    {
        int k=wtf(l, a[i]);
        dp[k]=a[i];
        l=max(l, k);
    }
    return l;
}
int main()
{
    while(~scanf("%d", &n))
    {
        memset(dp, 0, sizeof(dp));
        memset(a, 0, sizeof(a));
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);
        int ans=DP();
        printf("%d\n", ans);
    }
    return 0;
}