poj 2533 & poj 1631 Longest Ordered Subsequence( LIS果题 )

题目链接：

POJ 2533：http://poj.org/problem?id=2533

POJ 1631：http://poj.org/problem?id=1631

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i1, a_i2, ..., a_iK), where 1 <= i₁ < i₂ < ... < i_K <= 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

71 7 3 5 9 4 8

Sample Output

4

Source

Northeastern Europe 2002, Far-Eastern Subregion

LIS果题。

代码如下：

#include <cstdio>#include <iostream>#include <algorithm>int N;int ans;int a[1017], dp[1017];int  bin(int len, int tem){    int l = 1, r = len;    while(l <= r)    {        int mid = (l+r)/2;        if(tem > dp[mid])            l = mid+1;        else            r = mid-1;    }    return l;}int LIS(int *b){    dp[1] = a[1];    ans = 1;    int k;    for(int i = 2; i <= N; i++)    {        if(a[i] < dp[1])            k = 1;        else if(a[i] > dp[ans])            k = ++ans;        else            k = bin(ans,a[i]);        dp[k] = a[i];    }    return ans;}int main(){    while(~scanf("%d",&N))    {        for(int i = 1; i <= N; i++)        {            scanf("%d",&a[i]);        }        LIS(a);        printf("%d\n",ans);    }    return 0;}

POJ 1631 和这题基本上一样，只需改一下数组大小，叫一个多测试案例输入的循环即可！

这里也贴一下代码：

#include <cstdio>#include <iostream>#include <algorithm>int N;int ans;int a[40017], dp[40017];int  bin(int len, int tem){    int l = 1, r = len;    while(l <= r)    {        int mid = (l+r)/2;        if(tem > dp[mid])            l = mid+1;        else            r = mid-1;    }    return l;}int LIS(int *b){    dp[1] = a[1];    ans = 1;    int k;    for(int i = 2; i <= N; i++)    {        if(a[i] < dp[1])            k = 1;        else if(a[i] > dp[ans])            k = ++ans;        else            k = bin(ans,a[i]);        dp[k] = a[i];    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&N);        for(int i = 1; i <= N; i++)        {            scanf("%d",&a[i]);        }        LIS(a);        printf("%d\n",ans);    }    return 0;}