**【杭电1025】LIS 二分

 Constructing Roads In JGShining's Kingdom

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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.
两个for直接超时
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
	int x,y;
}a[500001];
int num[500001];
int main()
{
	int n;
	int k=1;
	while(~scanf("%lld",&n))
	{
	int  maxm=0;
	    for(int i=1;i<=n;i++)
	    {
		scanf("%d%d",&a[i].x,&a[i].y);
			num[i]=1;
	    	for(int j=1;j<i;j++)
	    	{
	    		if(a[i].x>a[j].x&&a[i].y>a[j].y)
	    		if(num[i]<=num[j]+1)
				num[i]=num[j]+1;	
			}
			maxm=max(maxm,num[i]);
		}
		printf("Case %d:\n",k++);
		if(maxm==1)
		printf("My king, at most %d road can be built.\n",maxm);
		else
		printf("My king, at most %d roads can be built.\n",maxm)
	}
	return 0;
}

看了别人的，才知道要用二分。╭∩╮（︶︿︶）╭∩╮   

详解：http://blog.youkuaiyun.com/lishuhuakai/article/details/8168328       
<span style="font-size:14px;">#include<iostream>
using namespace std;

long d[500001],a[500001];  //d[i]用来存最长不降子序列 最后一个元素 
long find(long a,long left,long right)  //二分查找
{
	long mid;
	while(left<=right)
	{
		mid=(left+right)/2;
       if(a>d[mid])//偏小 
		   left=mid+1;
	   else
		   right=mid-1;
	}
	return left;
}
int main()
{
	long n,i,j,len,p,t,k=0;
	while(scanf("%d",&n)!=EOF)
	{
		k++;
		for(i=1;i<=n;i++)
		{scanf("%d%d",&j,&p);
		a[j]=p;}
		d[1]=a[1];
		len=1;//初始序列长度 
		for(i=2;i<=n;i++)   
		{
			t=find(a[i],1,len);   //找出a[i]的位置
			d[t]=a[i];            //更新d[t],使d[t]尽量是最小的,显而易见，dp[t]越小，后面的dp[k](k>t)更大的可能性就会越大 
			//为什么可以这样做？仔细观察，发现对于一个a[i]，要判断它是否是某个不降子序列的元素，我们只需要拿它和前面的dp[j](j<len)比较即可，
			//由于dp[j]记录的是到a[j]为止的序列的最长不降子序列的最后一个元素，如果a[i]>dp[j]并且a[i].b<dp[j+1],
			//那么我们自然要用a[i]更新dp[j+1](很明显，最长子序列长度为j+1),而如果a[i]比dp[j]的值都大，
			//这说明a[i]可以接在所有的dp[j]后面，自然我们选择最长的dp[len],因此，加入之后，len的长度要自加1  
            if(t>len)            //因为d[len]都是最大的
				len++;
		}
   if(len==1)
     printf("Case %d:\nMy king, at most %d road can be built.\n\n",k,len);  //单数
   else
     printf("Case %d:\nMy king, at most %d roads can be built.\n\n",k,len);  //复数
	}
	return 0;
}</span>