hdu1025 （最长上升子序列n*logn）

Problem Description

JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines.

Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource. 

With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.

Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II. 

The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones. 

But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.

For example, the roads in Figure I are forbidden.



In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^

 

Input

Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.

 

Output

For each test case, output the result in the form of sample. 
You should tell JGShining what's the maximal number of road(s) can be built. 

 

Sample Input

   
   
   
   

    
    
    
    2
1 2
2 1
3
1 2
2 3
3 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
My king, at most 1 road can be built.

Case 2:
My king, at most 2 roads can be built.


    
    
    
    
 
     
 
      Hint 
     
 Huge input, scanf is recommended. 
    
   
   
   
   

这题简直没想到是最长上升子序列，而且要将两个值一个是数组下标，一个是数组值！这题接触了一种新的dp方式，时间复杂度只有nlogn，具体dp原理参考以下网站：

http://www.felix021.com/blog/read.php?1587

n*logn模板（最长不下降）：

int d[1000005];

int upper_bound(int arr[], int s, int e, int key)
{
    int mid;
    if (arr[e] <= key)
        return e + 1;
    while (s < e)
    {
        mid = s + (e - s) / 2;
        if (arr[mid] <= key)
            s = mid + 1;
        else
            e = mid;
    }
    return s;
}
int LIS(int d[], int n)
{
    int i = 0, len = 1, *end = (int *)malloc(sizeof(int) * (n + 1));
    end[1] = d[0]; //初始化：长度为1的LIS末尾为d[0]
    for (i = 1; i < n; i++)
    {
        int pos = upper_bound(end, 1, len, d[i]); //找到插入位置
        end[pos] = d[i];
        if (len < pos) //按需要更新LIS长度
            len = pos;
    }
    return len;
}

附上此题代码（最长上升）：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[500005],b[500005];
int main()
{
	int n,o=1;
	while(scanf("%d",&n)!=EOF)
	{
        for(int i=0;i<n;i++)
        {
            int c,b;scanf("%d%d",&c,&b);
          //  if(c>b){int t=c;c=b;b=t;}
            a[c]=b;
        }
        int j=1;
        b[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            //cout<<j<<endl;

                int low=1,high=j,mid;
                while(low<=high)
                {
                    mid=(low+high)/2;
                    if(a[i]>b[mid])
                        low=mid+1;
                    else high=mid-1;
                }
               // int k=lower_bound(b,b+j,a[i])-b;//cout<<' '<<k<<endl;
                b[low]=a[i];
                if(low>j)j++;

           // cout<<j<<endl;
        }
        if(j==1)
        printf("Case %d:\nMy king, at most %d road can be built.\n\n",o,j);
        else
           printf("Case %d:\nMy king, at most %d roads can be built.\n\n",o,j);
        o++;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
	}
	return 0;
}