codeforces 340D Bubble Sort Graph(最长非递减子序列)

D. Bubble Sort Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a1, a2, ..., an 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, ..., an (1 ≤ ai ≤ n).

Output

Output a single integer — the answer to the problem.

Sample test(s)

input

3
3 1 2

output

2

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].

题目大意：注意看定义，就是不断的冒泡，然后排好序之后，停止。我们不断在加边，找一个最长的序列，使得他们之间没有边联系即可。最后可以找到规律，就是求最长非递减子序列。

      解题思路： 先看下朴素的o(n*n)会超时，o(n*log(n))可以过，需要dp加上二分。具体见代码.

      题目地址：D. Bubble Sort Graph

o(n*n)朴素方法：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
using namespace std;
int a[100005];
int dp[100005];

int main()
{
    int n,i,j,res;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=1;
        }
        //dp[i]=max(max(dp[j])+1,dp[i]);  a[i]>=a[j]
        res=0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<i;j++)
            {
                if(a[i]>=a[j]&&dp[j]+1>dp[i])
                    dp[i]=dp[j]+1;
            }
            if(dp[i]>res)
                res=dp[i];
        }

        printf("%d\n",res);
    }
    return 0;
}

AC代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
using namespace std;
int a[100005];
int b[100005];
int n;

int findd(int len,int p)   //二分查找<=p的位置
{
    int l,r,mid;
    l=1,r=len,mid=(l+r)>>1;
    while(l<=r)
    {
        if(p>b[mid]) l=mid+1;
        else if(p<b[mid]) r=mid-1;
        else return mid;
        mid=(l+r)>>1;
    }
    return l;
}

int LIS()
{
    int i,j,len=1;
    b[1]=a[0];
    for(i=1;i<n;i++)
    {
        j=findd(len,a[i]);
        b[j]=a[i];  //b[j]是指长度为j最大元素的值
        if(j>=len) len=j;
    }
    return len;
}

int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        printf("%d\n",LIS());
    }
    return 0;
}