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

标签：poj   lis   

题目链接：

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

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

Description

Input

Output

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

Source

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;
}

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

标签：poj   lis   

原文地址：http://blog.csdn.net/u012860063/article/details/39001081