poj1836-Alignment(LIS)

poj1836 原题链接:http://poj.org/problem?id=1836

Description

In the army, a platoon is composed by n soldiers. During the morning inspection, the soldiers are aligned in a straight line in front of the captain. The captain is not satisfied with the way his soldiers are aligned; it is true that the soldiers are aligned in order by their code number: 1 , 2 , 3 , . . . , n , but they are not aligned by their height. The captain asks some soldiers to get out of the line, as the soldiers that remain in the line, without changing their places, but getting closer, to form a new line, where each soldier can see by looking lengthwise the line at least one of the line's extremity (left or right). A soldier see an extremity if there isn't any soldiers with a higher or equal height than his height between him and that extremity. 

Write a program that, knowing the height of each soldier, determines the minimum number of soldiers which have to get out of line. 

Input

On the first line of the input is written the number of the soldiers n. On the second line is written a series of n floating numbers with at most 5 digits precision and separated by a space character. The k-th number from this line represents the height of the soldier who has the code k (1 <= k <= n).

There are some restrictions: 
• 2 <= n <= 1000 
• the height are floating numbers from the interval [0.5, 2.5] 

Output

The only line of output will contain the number of the soldiers who have to get out of the line.

Sample Input

8
1.86 1.86 1.30621 2 1.4 1 1.97 2.2

Sample Output

4

题目大意:输入一个数列,求删除最少的数，使得从序列中任取一个数h[i]，有h[1] ~ h[i]单增，或h[i] ~ h[n]单减;

思路:从1~n开始枚举i,枚举到k值时,k值作为dp1的终点,作为dp2的起点,然后分别对两边进行dp,h[1]~h[k]求最长不降子序列,h[k]~h[n]求最长不升子序列,len1,len2分别表示其长度,求出len1+len2的最大值后用n-len1-len2即为所求;

因为有一个枚举分界点的循环,如果用往常的o(n^2)的LIS算法那么复杂度为o(n^3),交上去会TLE,所以这里LIS的算法应该使用o(nlogn)的算法,详细内容我写在另外一篇博客:http://blog.youkuaiyun.com/acm513828825

很遗憾写了很久还是没有对,有dalao能够帮我看一下更好,以下为我的错误代码:

#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;
float arr[1001];
float ans[1001];
float arr2[1001];
int binary_search(int i,int len){
    int left,right,mid;
    left=0,right=len;
    while(left<right){
        mid = left+(right-left)/2;
        if(arr[mid]>=ans[i]) right=mid;
        else left=mid+1;
    }
    return left;
}
int binary_search_2(int i,int len){
    int left,right,mid;
    left=0,right=len;
    while(left<right){
        mid = left+(right-left)/2;
        if(arr[mid]<=ans[i]) right=mid;
        else left=mid+1;
    }
    return left;
}
int main()
{
    int n;
    cin>>n;
    int len1=0;
    int len2=0;
    for(int i=1;i<=n;i++)
    {
        cin>>ans[i];
    }
    int Max=1;
    for(int k=1;k<=n;k++)
    {
        arr[0]=-1;
        arr[1]=ans[1];
        len1=1;
        for(int i=1+1;i<=k;i++)
        {
            if(arr[len1]<ans[i])
            {
                len1++;
                arr[len1]=ans[i];
            }
            else
            {
                int pos=binary_search(i,len1);
                arr[pos]=ans[i];
            }
        }
        if(k<=n-1)
        {
            arr2[1]=ans[k+1];
        len2=1;
        arr2[0]=-1;
        for(int j=k+2;j<=n;j++)
        {
            if(arr[len2]>ans[j])
            {
                len2++;
                arr2[len2]=ans[j];
            }
            else
            {
                int pos2=binary_search_2(j,len2);
                arr2[pos2]=ans[j];
            }
        }}
        else len2=0;
        if(Max<len1+len2)
        {
            Max=len1+len2;
        }
    }
    cout<<n-Max<<endl;
}

这里为正确代码,摘自:http://blog.youkuaiyun.com/lyy289065406/article/details/6648129

//Memory Time   
//232K   391MS   
  
//O(n*logn)算法，注意LIS和LDS使用不同的二分法  
#include<iostream>  
using namespace std;  
const int inf=3;  
  
//ord[]为不降序列  
//二分法搜索digit，若str中存在digit，返回其下标  
//若不存在，返回str中比digit小的最大那个数的（下标+1）  
int binary_search_1(double ord[],double digit,int head,int length)  
{  
    int left=head,right=length;  
    int mid;  
    while(right!=left)  
    {  
        mid=(left+right)/2;  
        if(digit==ord[mid])  
            return mid;  
        else if(digit<ord[mid])  
            right=mid;  
        else  
            left=mid+1;  
    }  
    return left;  
}  
  
//ord[]为不升序列  
//二分法搜索digit，若str中存在digit，返回其下标  
//若不存在，返回str中比digit大的最小那个数的（下标+1）  
int binary_search_2(double ord[],double digit,int head,int length)  
{  
    int left=head,right=length;  
    int mid;  
    while(right!=left)  
    {  
        mid=(left+right)/2;  
        if(digit==ord[mid])  
            return mid;  
        else if(digit>ord[mid])  
            right=mid;  
        else  
            left=mid+1;  
    }  
    return left;  
}  
  
int main(int i,int j)  
{  
    int n;  //士兵数  
    while(cin>>n)  
    {  
        double* h=new double[n+1];  
  
        for(i=1;i<=n;i++)  
            cin>>h[i];  
  
        int max=0;    
        for(int m=1;m<=n;m++)  //对身高队列每一个值作为分界点，进行枚举  
        {  
            double* ord=new double[n+1];  
  
            /*Dp-(0~m)-LIS*/  
  
            ord[0]=-1;  //下界无穷小  
            int len_LIS=1;  
            for(i=1;i<=m;i++)  
            {  
                ord[len_LIS]=inf; //上界无穷大  
                j=binary_search_1(ord,h[i],0,len_LIS);  
                if(j==len_LIS)  //sq[i]大于ord最大（最后）的元素  
                    len_LIS++;  
                ord[j]=h[i];  
            }  
            len_LIS--; //减去ord[0]的长度1  
  
            /*Dp-(m+1~n)-LDS*/  
  
            ord[m]=inf;  //下界无穷大  
            int len_LDS=1;  
            for(i=m+1;i<=n;i++)  
            {  
                ord[m+len_LDS]=-1; //上界无穷小  
                j=binary_search_2(ord,h[i],m,m+len_LDS);  
                if(j==m+len_LDS)  //sq[i]大于ord最小（最后）的元素  
                    len_LDS++;  
                ord[j]=h[i];  
            }  
            len_LDS--;  //减去ord[m]的长度1  
  
            //max为对于当前m的 最长不升子序列LIS 和 最长不降子序列LDS 长度之和  
  
            if(max<len_LIS+len_LDS)  
                max=len_LIS+len_LDS;  
  
            delete ord;  
        }  
          
        cout<<n-max<<endl;  
  
        delete h;  
    }  
    return 0;  
}