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.
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 a single integer — the answer to the problem.
3 3 1 2
2
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].
#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;
}#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) //二分查詢<=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]是指長度為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;
}