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N - Longest Ordered Subsequence POJ 2533 (最长上升子序列 )

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N - Longest Ordered Subsequence
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2533

Description
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<string>
#include<cstdlib>
#include<cmath>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1e4+10;
const double inf=99999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod  10007
int a[maxn];
int dp[maxn];
void solve()
{
    int n;
    scanf("%d",&n);
    fill(dp,dp+n,inf);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=0;i<n;i++){
        *lower_bound(dp,dp+n,a[i])=a[i];
    }
    printf("%d\n",lower_bound(dp,dp+n,inf)-dp);
}

int main()
{
    solve();
    return 0;
}















N - Longest Ordered Subsequence POJ 2533 (最长上升子序列 )

原文:http://blog.csdn.net/u013167299/article/details/45438289

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