最长递增子序列（LIS）

一:

很容易想到的 DP的O(N^2)的复杂度

#pragma comment(linker,"/STACK:102400000,102400000")

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>

using namespace std;

#define clc(a, b) memset(a, b, sizeof(a))

const int inf = 0x3f;

const int INF = 0x3f3f3f3f;

const int maxn = 1000;

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



int LIS()

{

    int i, j, k = 0;

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

    {

        dp[i] = 1;

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

        {

            if(a[i] > a[j] && dp[i] < dp[j] + 1)

            {

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

            }

        }

        k = dp[i] > k ? dp[i] : k;

    }

    return k;

}

int main()

{

    while(~scanf("%d", &n))

    {

        clc(dp, 0);

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

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

        printf("LIS = ");

        printf("%d\n", LIS());

    }

}

二: 扩展升级版, 求定长的上升子序列个数

#pragma comment(linker,"/STACK:102400000,102400000")

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>

using namespace std;

#define clc(a, b) memset(a, b, sizeof(a))

const int inf = 0x3f;

const int INF = 0x3f3f3f3f;

const int maxn = 1000;

const int mod = 1000000007;

int n, m, a[maxn], dp[maxn][maxn], sum[maxn];

//sum 记录长度为i的子序列个数 dp[i][j] 记录从i开始长度为j的个数

int LIS()

{

    int i, j, k;

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

    {

        dp[i][1] = 1;

        sum[1] = (dp[i][1] + sum[1]) % mod;

        for(k = 2; k <= i; k++)

        {

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

            {

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

                {

                    dp[i][k] = (dp[i][k] + dp[j][k-1]) % mod;

                }

            }

            sum[k] = (sum[k] + dp[i][k]) % mod;

            //printf("i = %d, sum[%d] = %d\n", i, k, sum[k]);

        }

    }

    return 0;

}

int main()

{

    int q;

    while(~scanf("%d %d", &n, &q))//n 数组长度, q个询问

    {

        clc(dp, 0);

        clc(sum, 0);

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

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

        LIS();

        while(q--)

        {

            scanf("%d", &m);

            printf("%d\n", sum[m]);

        }

    }

    return 0;

}

三: DP + 二分法

链接: 原理+解释很详细http://www.felix021.com/blog/read.php?1587

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>



using namespace std;



const int inf = 0x3f;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5+5;



int lis[maxn], a[maxn], n;



int BinSearch(int len, int x)

{

    int left = 0, right = len - 1;

    while(left <= right)

    {

        int mid = (left + right) / 2;

        if(lis[mid] <= x)

        {

            left = mid + 1;

        }

        else 

        {

            right = mid - 1;

        }

    }

    return left;

}

int LIS()

{

    lis[0] = a[0];

    int len = 1;

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

    {

        if(a[i] > lis[len-1])

        {

            lis[len++] = a[i];

        }

        else

        {

            int pos = BinSearch(len, a[i]);

            lis[pos] = a[i];

        }

    }

    return len;

}

int main()

{

    while(~scanf("%d", &n))

    {

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

        {

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

        }



        int ans = LIS();

        printf("%d\n", ans);

    }

}