HDU-1025-Constructing Roads In JGShining's Kingdom【LIS】【二分】

題目連結：點選打開連結

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23245    Accepted Submission(s): 6645

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.

   
    
        

看了别人的思路，才知道可以用二分找：

解題報告：就是求最長上升子序列，一開始用的普通辦法求的！直接TEL；就在網上找了一個時間複雜度為O(nlogn)的算法，其算法思想為：（網上找的）

假設要尋找最長上升子序列的序列是a[n]，然後尋找到的遞增子序列放入到數組b中。

（1）當周遊到數組a的第一個元素的時候，就将這個元素放入到b數組中，以後周遊到的元素都和已經放入到b數組中的元素進行比較；

（2）如果比b數組中的每個元素都大，則将該元素插入到b數組的最後一個元素，并且b數組的長度要加1；

（3）如果比b數組中最後一個元素小，就要運用二分法進行查找，查找出第一個比該元素大的最小的元素，然後将其替換。

在這個過程中，隻重複執行這兩步就可以了，最後b數組的長度就是最長的上升子序列長度。例如：如該數列為：

5 9 4 1 3 7 6 7

那麼：

5 //加入

5 9 //加入

4 9 //用4代替了5

1 9 //用1代替4

1 3 //用3代替9

1 3 7 //加入

1 3 6 //用6代替7

1 3 6 7 //加入

最後b中元素的個數就是最長遞增子序列的大小，即4。

要注意的是最後數組裡的元素并不就一定是所求的序列，

例如如果輸入 2 5 1

那麼最後得到的數組應該是 1 5

而實際上要求的序列是 2 5

#include<cstdio>
#include<algorithm>
#include<cstring>
const int MAX=1e6+10;
int n;
int city[MAX];
int sort_city[MAX];
//用二分查找的方法找到一個位置，使得 x>sort_city[i-1] 并且 x<sort_city[i],并用 x 代替sort_city[i]
int Search(int x,int left,int right)
{
	int mid;
	while(left<=right)
	{
		mid=(left+right)>>1;
		if(x>sort_city[mid])
			left=mid+1;
		else
			right=mid-1;
	}
	return left;
}
int DP()
{
	int len=1;
	sort_city[len]=city[1];
	for(int i=2;i<=n;i++)
	{
		if(city[i]>sort_city[len])//如果 city[i]比 sort_city[]數組中最大還大,就直接插入到後面即可
		{
			len++;
			sort_city[len]=city[i];
		}
		else//用二分的方法在 sort_city[]數組中找出第一個比 city[i]大的位置,并且讓 city[i]替代這個位置
		{
			int id=Search(city[i],1,len);
			sort_city[id]=city[i];
		}
	}
	return len;
}
int main()
{
	int text=0;
	while(~scanf("%d",&n))
	{
		int a,b;
		for(int i=1;i<=n;i++)
		{
			scanf("%d %d",&a,&b);
			city[a]=b;
		}
		printf("Case %d:\n",++text);
		if(DP()==1) // 這裡注意下輸出格式（ road 還要複數形式，看來出題人英語學得不錯） 
			printf("My king, at most 1 road can be built.\n\n");
		else
			printf("My king, at most %d roads can be built.\n\n",DP());
	}
	return 0;
}