Codeforces Round #198 (Div. 2) D. Bubble Sort Graph （转化为最长非降子序列）

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].

思路：

问题等价于找一个最长非降子序列。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 100005
using namespace std;

int n,m,ans,cnt;
int a[maxn],dp[maxn];

void solve()
{
int i,j,pos;
dp[0]=0;
cnt=0;
for(i=1; i<=n; i++)
{
if(a[i]>=dp[cnt]) dp[++cnt]=a[i];
else
{
pos=upper_bound(dp,dp+cnt+1,a[i])-dp;  // 找到>a[i]的第一次出现的位置
printf("i:%d pos:%d\n",i,pos);
dp[pos]=a[i];low
}
}
}
int main()
{
int i,j;
while(~scanf("%d",&n))
{
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
solve();
printf("%d\n",cnt); // 长度即为cnt 但序列不是dp保存的序列 要输出序列的话应在更新ant时记录序列
}
return 0;
}
/*
7
2 3 3 5 3 2 4
*/