Constructing Roads In JGShining's Kingdom HDU - 1025(二分优化)

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最长上升子序列

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本文介绍了一个基于二分查找优化的最长上升子序列算法，用于解决资源丰富城市与资源匮乏城市之间的资源配对问题，确保道路不交叉且实现最大数量的配对。

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.

在这里插入图片描述

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.

题意

在富裕城市与贫困城市之间修建道路，不能交叉且一对一，最多能修建几条。

分析

因为保证了一条线上的城市有序，所以求另一条线上最长上升子序列即可，直接求解n*n肯定会tle，所以要用到二分来优化。
二分查找：每次与中间的值比较，如果小了，就取上区间，大了就取下区间。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

#define maxn 500500
int a[maxn]; 
int dp[maxn];
int find(int x,int low,int high)
{
	while(low<=high)
	{
		int mid=(high+low)>>1;
		if(dp[mid]<a[x])
		{
			low=mid+1;
		}
		else high=mid-1;
	}
	return low;
}
int main()
{
	int n,cnt=0;
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			a[u]=v;
		}
		memset(dp,0,sizeof(dp));
		int len=0;//记录最长结果
		for(int i=1;i<=n;i++){
			if(a[i]>dp[len]){
				dp[++len]=a[i];
			}
			else {
				dp[find(i,1,len)]=a[i];//每次遇到一个小于此值的就在前面找一个将其替换，以此来满足最长，因为相对大小不影响结果
			}
		}
		if(len==1)
		printf("Case %d:\nMy king, at most %d road can be built.\n",++cnt,len);
		else printf("Case %d:\nMy king, at most %d roads can be built.\n",++cnt,len);

	}
	return 0;
}