POJ_2533 Longest Ordered Subsequence【DP】【最长递增子序列】

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 58448 Accepted: 26207

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

题意

给出一个数组，求数组中的最长递增子序列的长度

思路一

我们可以用两个数组，第一个数组为原数组，第二个数组为原数组经过排序加去重（如果是非下降子序列就不需要去重），然后求两个数组的最长公共子序列就可以了。

AC代码一

#include <iostream>          //转化为求LCS

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <deque>

#include <vector>

#include <queue>

#include <string>

#include <cstring>

#include <map>

#include <stack>

#include <set>

#include <cstdlib>

#include <ctype.h>

#include <numeric>

#include <sstream>

using namespace std;

typedef long long LL;

const double PI  = 3.14159265358979323846264338327;

const double E   = 2.718281828459;

const double eps = 1e-6;

const int MAXN   = 0x3f3f3f3f;

const int MINN   = 0xc0c0c0c0;

const int maxn   = 1e3 + 5;

const int MOD    = 1e9 + 7;

int a[maxn], b[maxn], dp[maxn][maxn];

int main()

{

    int n;

    cin >> n;

    int i, j;

    for (i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

        b[i] = a[i];

    }

    sort(b, b + n);

    int dre = unique(b, b + n) - b;    //需要去重  因为是最长上升的 如果是非下降，那么不需要去重

    memset(dp, 0, sizeof(dp));

    for (i = 0; i < dre; i++)

    {

        if (a[0] == b[i])

            dp[0][i] = 1;

        else if (i)

            dp[0][i] = dp[0][i - 1];

    }

    for (i = 0; i < n; i++)

    {

        if (b[0] == a[i])

            dp[i][0] = 1;

        else if (i)

            dp[i][0] = dp[i - 1][0];

    }

    for (i = 1; i < n; i++)

    {

        for (j = 1; j < dre; j++)

        {

            if (a[i] == b[j])

                dp[i][j] = dp[i - 1][j - 1] + 1;

            else

                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

        }

    }

    cout << dp[n - 1][dre - 1] << endl;

}

思路二

如果一个数小于它前面的一个数，那么到这个数为止的最长上升子序列就是前面那个数的最长上升子序列 + 1 然后每个数往前扫一遍就可以了

AC代码二

#include <iostream>    //DP

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <deque>

#include <vector>

#include <queue>

#include <string>

#include <cstring>

#include <map>

#include <stack>

#include <set>

#include <cstdlib>

#include <ctype.h>

#include <numeric>

#include <sstream>

using namespace std;

typedef long long LL;

const double PI  = 3.14159265358979323846264338327;

const double E   = 2.718281828459;

const double eps = 1e-6;

const int MAXN   = 0x3f3f3f3f;

const int MINN   = 0xc0c0c0c0;

const int maxn   = 1e3 + 5;

const int MOD    = 1e9 + 7;

int arr[maxn], dp[maxn];

int main()

{

    int n;

    cin >> n;

    int i, j;

    for (i = 0; i < n; i++)

        scanf("%d", &arr[i]);

    memset(dp, 0, sizeof(dp));

    int ans = 1;

    for (i = 0; i < n; i++)

    {

        dp[i] = 1;

        for (j = i - 1; j >= 0; j--)

        {

            if (arr[i] > arr[j])

                dp[i] = max(dp[i], dp[j] + 1);

        }

        if (dp[i] > ans)

            ans = dp[i];

    }

    cout << ans << endl;

}